{"id":31534,"date":"2026-05-20T17:10:11","date_gmt":"2026-05-20T11:40:11","guid":{"rendered":"https:\/\/mycbseguide.com\/blog\/?p=31534"},"modified":"2026-05-20T17:10:52","modified_gmt":"2026-05-20T11:40:52","slug":"how-forces-affect-motion-ncert-solutions-class-9-science-exploration","status":"publish","type":"post","link":"https:\/\/mycbseguide.com\/blog\/how-forces-affect-motion-ncert-solutions-class-9-science-exploration\/","title":{"rendered":"How Forces Affect Motion &#8211; NCERT Solutions Class 9 Science Exploration"},"content":{"rendered":"\n<p>How Forces Affect Motion &#8211; NCERT Solutions Class 9 Science Exploration includes all the questions with solutions given in NCERT Class 9 <a href=\"https:\/\/ncert.nic.in\/textbook.php?iesc1=0-13\">Science Exploration<\/a> textbook.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">NCERT Solutions Class 9<\/h2>\n\n\n<a class=\"mks_button mks_button_small rounded\" href=\"https:\/\/mycbseguide.com\/blog\/category\/ncert-solutions\/ncert-solutions-class-9\/ncert-solutions-class-9-english-kaveri\/\" target=\"_self\" style=\"color: #FFFFFF; background-color: #0066bf;\" >English Kaveri<\/a>\n\n\n<a class=\"mks_button mks_button_small rounded\" href=\"https:\/\/mycbseguide.com\/blog\/category\/ncert-solutions\/ncert-solutions-class-9\/ncert-solutions-class-9-hindi-ganga\/\" target=\"_self\" style=\"color: #FFFFFF; background-color: #0066bf;\" >Hindi Ganga<\/a>\n\n\n<a class=\"mks_button mks_button_small rounded\" href=\"https:\/\/mycbseguide.com\/blog\/category\/ncert-solutions\/ncert-solutions-class-9\/ncert-solutions-class-9-sanskrit-sharada\/\" target=\"_self\" style=\"color: #FFFFFF; background-color: #0066bf;\" >Sanskrit Sharada<\/a>\n\n\n<a class=\"mks_button mks_button_small rounded\" href=\"https:\/\/mycbseguide.com\/blog\/category\/ncert-solutions\/ncert-solutions-class-9\/ncert-solutions-class-9-maths-ganita-manjari\/\" target=\"_self\" style=\"color: #FFFFFF; background-color: #0066bf;\" >Maths Ganita Manjari<\/a>\n\n\n<a class=\"mks_button mks_button_small rounded\" href=\"https:\/\/mycbseguide.com\/blog\/category\/ncert-solutions\/ncert-solutions-class-9\/ncert-solutions-class-9-science-exploration\/\" target=\"_self\" style=\"color: #FFFFFF; background-color: #0066bf;\" >Science Exploration<\/a>\n\n\n<a class=\"mks_button mks_button_small rounded\" href=\"https:\/\/mycbseguide.com\/blog\/category\/ncert-solutions\/ncert-solutions-class-9\/ncert-solutions-class-9-social-understanding-society\/\" target=\"_self\" style=\"color: #FFFFFF; background-color: #0066bf;\" >Social Understanding Society<\/a>\n\n\n\n<h2 class=\"wp-block-heading\">How Forces Affect Motion \u2013 NCERT Solutions<\/h2>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.1: Why does a canoe move forward when the canoeist pushes water backwards with their paddle and why does it move faster when they push harder?<\/strong><\/p>\n\n\n\n<p>Solution: When the canoeist pushes water backward with the paddle, the water exerts an equal and opposite force on the canoe, pushing it forward. This is why the canoe moves ahead. When the canoeist pushes the water harder, a larger force is applied backward, so the water gives a larger forward reaction force, making the canoe accelerate more and move faster.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.2: Suppose the same canoeist uses the same paddle force in two different canoes, one empty and one carrying another passenger. In which case will the canoe move faster?<\/strong><\/p>\n\n\n\n<p>Solution: The canoe with only the canoeist (empty canoe) will move faster.<br>Because the same force produces more acceleration when the mass is smaller. The canoe with an extra passenger has greater mass, so it accelerates less and moves slower.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.3: A weightlifter lifts a barbell. List two forces that are acting on the barbell. Are these forces balanced if the weightlifter keeps the barbell steady?<\/strong><br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1776947018-f3v2vx.jpg\"><\/p>\n\n\n\n<p>Solution: Forces acting on the barbell:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Downward force due to gravity (weight of barbell)<\/li>\n\n\n\n<li>Upward force applied by the weightlifter<\/li>\n<\/ol>\n\n\n\n<p>Yes. When the barbell is held steady, both forces are equal in magnitude and opposite in direction, so the net force is zero and the forces are balanced.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.4: Two players R and S are participating in an arm-wrestling match. At the instant, when the arms tilt to the front direction (out of the page towards you), are the forces exerted by the players balanced? If not, which player exerted the larger force?<\/strong><br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1776947243-85xuev.jpg\"><\/p>\n\n\n\n<p>Solution: No, the forces are not balanced at that instant because the arms are moving (tilting), so there is a net force.<\/p>\n\n\n\n<p>The player whose arm is moving forward (towards you \/ out of the page) is exerting the larger force and is winning the match.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.5: An object is moving with a constant velocity. Is there a net force acting upon it?<\/strong><\/p>\n\n\n\n<p>Solution: No, there is no net force acting on the object.<br>According to Newton&#8217;s First Law of Motion, an object moving with constant velocity continues in uniform motion unless acted upon by an external unbalanced force. Since the velocity is constant, the acceleration is zero. From Newton&#8217;s Second Law, {tex}F=m a{\/tex}, if {tex}a=0{\/tex}, then net force {tex}F=0{\/tex}.<\/p>\n\n\n\n<p>Therefore, no net force acts on the object.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.6: Suppose, no net force is acting on an object. Which of the following situations are possible?<\/strong><\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Object remains at rest if at rest.<\/li>\n\n\n\n<li>Object keeps moving with a constant velocity if already moving.<\/li>\n\n\n\n<li>Object is moving with a constant acceleration.<\/li>\n<\/ol>\n\n\n\n<p>Solution: If no net force is acting on an object, then its acceleration is zero (Newton\u2019s First Law of Motion).<\/p>\n\n\n\n<p>Now, analysing the given situations:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li><strong>Object remains at rest if at rest:<\/strong><br>Yes, this is possible because without any net force, an object at rest continues to remain at rest.<\/li>\n\n\n\n<li><strong>Object keeps moving with a constant velocity if already moving:<\/strong><br>Yes, this is also possible because in absence of net force, a moving object continues with uniform velocity.<\/li>\n\n\n\n<li><strong>Object is moving with a constant acceleration:<\/strong><br>No, this is not possible because constant acceleration requires a net force to act on the object.<\/li>\n<\/ol>\n\n\n\n<p><strong>Conclusion:<\/strong> Situations (i) and (ii) are possible, but (iii) is not possible.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.7: In the real world, it is difficult to find a situation where no forces are acting on an object. But by applying additional forces, a condition can be achieved where the net force on the object is zero. Explain with the help of an example.<\/strong><\/p>\n\n\n\n<p>Solution: In real life, multiple forces usually act on an object. However, if these forces balance each other such that their resultant (net force) is zero, the object behaves as if no force is acting on it.<\/p>\n\n\n\n<p><strong>Example:<\/strong><br>When a book is kept on a table, two forces act on it:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The weight of the book acting downward due to gravity.<\/li>\n\n\n\n<li>The normal reaction force exerted by the table acting upward.<\/li>\n<\/ul>\n\n\n\n<p>These two forces are equal in magnitude and opposite in direction, so they cancel each other. Hence, the net force on the book is zero, and it remains at rest.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.8: A toy car of mass 100 g is moving with a constant velocity of 0.5 m s<sup>-1<\/sup>. What is the net force acting on the toy car?<\/strong><\/p>\n\n\n\n<p>Solution: Given:<br>Mass of toy car, {tex}m=100 {~g}=0.1 {~kg}{\/tex}<br>Velocity, {tex}v=0.5 m s^{-1}{\/tex} (constant)<br>Since the velocity is constant, acceleration {tex}a=0{\/tex}.<br>Using Newton&#8217;s Second Law:<\/p>\n\n\n\n<p>{tex} F=m a=0.1 \\times 0=0{\/tex}<\/p>\n\n\n\n<p>Net force acting on the toy car {tex}=0 {~N}{\/tex}.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.9: Two children of different masses are sitting on identical swings. To impart identical initial acceleration, for which child would you require to apply a larger force? Explain why.<\/strong><\/p>\n\n\n\n<p>Solution: According to Newton&#8217;s Second Law of Motion, {tex}F=m a{\/tex}.<br>For identical initial acceleration {tex}a{\/tex}, the force required is directly proportional to the mass {tex}m{\/tex} of the child.<\/p>\n\n\n\n<p>Therefore, the child with greater mass requires a larger force to achieve the same acceleration.<\/p>\n\n\n\n<p><strong>Conclusion:<\/strong> A larger force must be applied to the heavier child.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.10: How are glass items packed for transportation using a bubble wrap or hay protected from damage?<\/strong><\/p>\n\n\n\n<p>Solution: Glass items are packed with bubble wrap or hay to increase the time over which the momentum of the glass becomes zero during a collision.<\/p>\n\n\n\n<p>According to the relation {tex}F=\\frac{\\Delta p}{\\Delta t}{\/tex}, if the time of impact {tex}\\Delta t{\/tex} increases, the force {tex}F{\/tex} decreases.<\/p>\n\n\n\n<p>Bubble wrap or hay acts as a cushion, reducing the force of impact. Hence, the glass items are protected from damage.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.11: Why does a fireperson sometimes struggle when holding the pipe issuing water?<\/strong><\/p>\n\n\n\n<p>Solution: When water comes out of the pipe with high velocity, it carries a large momentum. According to Newton\u2019s Third Law of Motion, the water exerts an equal and opposite reaction force on the pipe.<\/p>\n\n\n\n<p>This reaction force acts backward on the fireperson holding the pipe, making it difficult to control.<\/p>\n\n\n\n<p>Hence, the fireperson sometimes struggles due to the backward force exerted by the fast-flowing water.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.12: Suppose a spacecraft is moving in a region of space where the gravitational force acting upon it is negligible. Suggest how can it change its velocity.<\/strong><\/p>\n\n\n\n<p>Solution: In the absence of external forces, a spacecraft will continue to move with constant velocity (Newton\u2019s First Law of Motion).<\/p>\n\n\n\n<p>To change its velocity, the spacecraft can <strong>expel gases in the opposite direction<\/strong> using its engines. According to Newton\u2019s Third Law of Motion, the expelled gases exert an equal and opposite reaction force on the spacecraft, causing it to accelerate.<\/p>\n\n\n\n<p>Thus, by firing its thrusters, the spacecraft can change its velocity even in space.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.13: Two forces of 10 N and 6 N are acting on a block lying on the table as shown in the figure. What is the magnitude and the direction of the net force acting on the block in each case?<\/strong><br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1776946711-efk9vp.jpg\"><\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Net force = 10 N + 6 N = 16 N, acting towards the right side.<\/li>\n\n\n\n<li>Net force = 10 N &#8211; 6 N = 4 N, acting towards the right side.<\/li>\n\n\n\n<li>Net force = 10 N &#8211; 6 N = 4 N, acting towards the left side.<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.14: A person is exerting a force on a moving box in the forward direction which is equal to the force of friction acting between the bottom surface of the box and the floor. Will the box continue moving or will it come to rest after some time?<\/strong><\/p>\n\n\n\n<p>Solution: The force of friction will be acting on the box in the backward direction. The two forces acting on the box are equal and opposite, and thus, they balance each other. The net force acting on the box is zero and as per the Newton\u2019s first law of motion, the box will continue moving with constant velocity.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.15: Draw (i) position-time, and (ii) velocity-time graphs for an object on which no net force is acting.<\/strong><\/p>\n\n\n\n<p>Solution: When no net force is acting on an object, there are two possibilities\u2009-\u2009either the object is at rest or the object is moving with a constant velocity. If the object is at rest, its position will not change with time and its position-time graph is as shown in Figure\u00a01a. Its velocity will remain zero and its velocity-time graph is shown in Figure 1b. If the object is moving with constant velocity and no net force is acting upon it, it will continue moving with the same velocity and the velocity-time graph is shown in Figure 2b. The position-time graph will be a straight line as shown in Figure\u00a02a.<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1777270452-mhphet.jpg\"><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.16: A weightlifter is holding a barbell with mass of 10 kg fixed on each side of the bar. The mass of the bar itself is 10 kg. How much force is she applying to keep the barbell steady?<\/strong><br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1776947018-f3v2vx.jpg\"><\/p>\n\n\n\n<p>Solution: The total mass of the barbell is 30 kg. The gravitational force, due to the Earth, acting on the barbell in the downward direction is (using Eq. F = mg),<\/p>\n\n\n\n<p>{tex} F=m g=30 {~kg} \\times 9.8 {~m} {s}^{-2}=294 {~N}{\/tex}<\/p>\n\n\n\n<p>To keep the barbell steady, the weightlifter has to apply an equal force in the opposite direction. So, she is applying 294 N in the upward direction.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.17: A student is trying to push a stationary block of 25 kg on a horizontal floor. The maximum force of friction opposing this motion is 50 N. Determine the displacement of the block in 2 seconds if Rahul pushes it with a constant force of (i) 50 N and (ii) 55 N in the forward direction.<\/strong><\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>The force applied by the student is equal to the opposing force of friction. Thus, the two forces are balanced and the net force acting on the block is zero. So, the block will remain stationary.<\/li>\n\n\n\n<li>The net force on the block is {tex}55 {~N}-50 {~N}=5 {~N}{\/tex}.<br>The mass of the block is 25 kg. Using the Newton&#8217;s second law of motion, the acceleration of the block is<br>{tex} a=\\frac{F}{m}=\\frac{5 {~N}}{25 {~kg}}=\\frac{5 {~kg} {~m} {~s}^{-2}}{25 {~kg}}=0.2 {~m} {~s}^{-2}{\/tex}<br>Using the kinematic equation, the displacement of the block in 2 seconds is {tex}s=u t+\\frac{1}{2} a t^2=\\left[0 {~m} {~s}^{-1} \\times 2 {~s}\\right]{\/tex}\u00a0{tex}+\\left[\\frac{1}{2} \\times 0.2 {~m} {~s}^{-2} \\times(2 {~s})^2\\right]=0.4 {~m}{\/tex} in the forward direction.<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.18: A sports car of mass 1500 kg is moving towards the east and its velocity-time graph is shown in the figure.<\/strong><br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1777004075-ch4ccd.jpg\"><br>Calculate the force acting on the car during<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>0 s to 5 s<\/li>\n\n\n\n<li>5 s to 10 s<\/li>\n\n\n\n<li>10 s to 15 s<\/li>\n<\/ol>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>During 0 s to 5 s:<br>As the velocity-time graph is a straight line inclined to the time axis, it indicates that the sports car is moving with a constant acceleration during this time interval with {tex}u=0 {~m} {s}^{-1}, v=10 {~m} {s}^{-1}{\/tex} and {tex}t=5 {~s}{\/tex}.<br>Using the kinematic equation {tex}v=u+a t{\/tex}, we can find the acceleration, {tex}10 {~m} {s}^{-1}=0 {~m} {s}^{-1}+(a \\times 5 {~s}){\/tex}<br>{tex}a=2 {~m} {s}^{-2}{\/tex} Now, using the Newton&#8217;s second law of the motion {tex}F=m a{\/tex}, we can find the force acting on the sports car as, {tex} F=1500 {~kg} \\times 2 {~m} {~s}^{-2}=3000 {~N} \\text { acting towards the east. }{\/tex}<\/li>\n\n\n\n<li>During 5 s to 10 s:<br>As the velocity-time graph is a straight line parallel to the time axis, it indicates that the sports car is moving with a constant velocity. Hence, no force is acting on the sports car.<\/li>\n\n\n\n<li>During 10 s to 15 s: As the velocity-time graph is a straight line inclined to the time axis, it indicates that the sports car is moving with a constant acceleration in this time interval with {tex}u=10 {~m} {~s}^{-1}, v=0 {~m} {~s}^{-1}{\/tex} and {tex}t=5 {~s}{\/tex}. Using the kinematic equation {tex}v=u+a t{\/tex}, we can find the acceleration, {tex}0 {~m} {~s}^{-1}=10 {~m} {~s}^{-1}+(a \\times 5 {~s}){\/tex}<br>{tex}a=-2 {~m} {~s}^{-2}{\/tex} Now, using the Newton&#8217;s second law of the motion {tex}F=m a{\/tex}, we can find the force acting on the sports car as, {tex} F=1500 {~kg} \\times\\left(-2 {~m} {~s}^{-2}\\right)=-3000 {~N}{\/tex} The negative sign shows that the force is acting in a direction opposite to the direction of motion, that is towards the west.<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.19: As shown in the figure, the Earth and the fruit apply equal and opposite gravitational forces on each other. Then why does the fruit move towards the Earth while the Earth doesn\u2019t seem to move towards the fruit?<\/strong><br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1777005038-695sq7.jpg\"><\/p>\n\n\n\n<p>Solution: Though the forces acting on both the Earth and the fruit are equal in magnitude; the mass of the Earth is so large (as compared to the fruit) that the acceleration of the Earth caused by the force is extremely small (as per {tex}a=\\frac{F}{m}{\/tex}). Thus, its effect on the Earth is too small to be noticed.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.20: When a 0.1 kg bullet is fired from a 5 kg gun with a force of 2 N, the gun recoils. What are the magnitudes of initial accelerations of the bullet and the gun?<\/strong><\/p>\n\n\n\n<p>Solution: From the Newton&#8217;s third law of motion, the recoil force on the gun is also 2 N.<\/p>\n\n\n\n<p>From the Newton&#8217;s second law of motion, the initial magnitudes of acceleration of gun<\/p>\n\n\n\n<p>{tex} =\\frac{\\text { force }}{\\text { mass of gun }}=\\frac{2 {~N}}{5 {~kg}}=0.4 {~m} {~s}^{-2}{\/tex}<\/p>\n\n\n\n<p>While the initial acceleration of bullet<\/p>\n\n\n\n<p>{tex} =\\frac{\\text { force }}{\\text { mass of bullet }}=\\frac{2 {~N}}{0.1 {~kg}}=20 {~m} {~s}^{-2}{\/tex}<\/p>\n\n\n\n<p>Even though the pair of forces are equal in magnitude, the magnitudes of accelerations are not equal because their masses are different.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.21: Using a horizontal force F, a table is moved across the floor at a constant velocity. How much is the frictional force exerted by the floor on the table?<\/strong><\/p>\n\n\n\n<p>Solution: When the table moves with constant velocity, its acceleration is zero.<br>Hence, the net force acting on the table is zero (Newton&#8217;s First Law of Motion).<\/p>\n\n\n\n<p>Therefore, the applied force {tex}F{\/tex} is balanced by the frictional force.<\/p>\n\n\n\n<p>{tex} \\text { Frictional force }=F{\/tex}<\/p>\n\n\n\n<p><strong>Conclusion: <\/strong>The frictional force exerted by the floor on the table is equal in magnitude and opposite in direction to the applied force {tex}F{\/tex}.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.22: For a ball moving on a smooth frictionless surface, choose the appropriate option that will make the following statements physically correct.<\/strong><\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>If no net force is applied on the ball, the velocity of the ball will remain the same\/increase\/decrease.<\/li>\n\n\n\n<li>If a net force is applied on the ball in the direction of its motion, the magnitude of the velocity of the ball will remain the same\/ increase\/decrease.<\/li>\n\n\n\n<li>If a net force is applied on the ball in a direction opposite to the direction of its motion, the magnitude of the velocity of the ball will remain the same\/increase\/decrease.<\/li>\n<\/ol>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li><strong>If no net force is applied on the ball:<\/strong><br>The velocity of the ball will <strong>remain the same<\/strong>.<br>(An object continues in uniform motion in absence of net force.)<\/li>\n\n\n\n<li><strong>If a net force is applied in the direction of motion:<\/strong><br>The magnitude of velocity will <strong>increase<\/strong>.<br>(Force produces acceleration in the same direction.)<\/li>\n\n\n\n<li><strong>If a net force is applied opposite to the direction of motion:<\/strong><br>The magnitude of velocity will <strong>decrease<\/strong>.<br>(Force produces acceleration opposite to motion, slowing it down.)<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.23: Two blocks P and Q on a smooth horizontal surface are shown in the figure a and figure b. Two forces of magnitudes 4 N and 5 N are acting in opposite directions on block P, while block Q is moving with a constant velocity.<\/strong><br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1777006069-8weh23.jpg\"><br>Which of the following statement is correct?<\/p>\n\n\n\n<p>Options:<br>(1) P experiences a net force and Q does not experience a net force. \u2705<br>(2) P does not experience a net force and Q experiences a net force.<br>(3) Both P and Q experience a net force.<br>(4) Neither P nor Q experiences a net force.<\/p>\n\n\n\n<p>Explanation:<\/p>\n\n\n\n<p>For block P:<br>Two forces of 4 N and 5 N act in opposite directions.<br>Net force {tex}=5-4=1 {~N}(\\neq 0){\/tex}<br>So, P experiences a net (unbalanced) force.<\/p>\n\n\n\n<p>For block Q:<br>It is moving with constant velocity, so acceleration is zero.<br>Hence, net force on {tex}{Q}=0{\/tex} (balanced forces).<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.24: While practising for the snake boat race (Vallam Kali in Kerala), 100 oarsmen are rowing a boat together. Out of these, 95 row backwards to propel the boat forward. But by mistake, 5 oarsmen row in the opposite direction. If each oarsman applies a horizontal force of 200 N, what is the net force on the snake boat? (Ignore drag forces, air friction, etc.)<\/strong><\/p>\n\n\n\n<p>Solution: Given:<br>Force applied by each oarsman {tex}=200 {~N}{\/tex}<br>Number of oarsmen rowing forward {tex}=95{\/tex}<br>Number of oarsmen rowing opposite {tex}=5{\/tex}<br>Total forward force {tex}=95 \\times 200=19000 {~N}{\/tex}<br>Total backward force {tex}=5 \\times 200=1000 {~N}{\/tex}<br>Net force {tex}=19000-1000=18000 {~N}{\/tex}<\/p>\n\n\n\n<p>Net force on the boat {tex}=18000 {~N}{\/tex} (in forward direction).<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.25: When a net force acts on an object, we observe that the object accelerates:<\/strong><\/p>\n\n\n\n<p>Options:<br>(1) opposite to the direction of force, with acceleration proportional to the force acting on the object.<br>(2) opposite to the direction of force, with acceleration proportional to the mass of the object.<br>(3) in the direction of force, with acceleration inversely proportional to the force acting on the object.<br>(4) in the direction of force, with acceleration proportional to the force acting on the object. \u2705<\/p>\n\n\n\n<p>Explanation: According to Newton\u2019s Second Law, acceleration produced in an object is:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Directly proportional to the applied force, and<\/li>\n\n\n\n<li>In the direction of the force.<\/li>\n<\/ul>\n\n\n\n<p>Thus, the correct statement is:<\/p>\n\n\n\n<p>in the direction of force, with acceleration proportional to the force acting on the object.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.26: The position-time graph for four objects A, B, C and D moving along a straight line are given in the figure. A net force acts on:<\/strong><br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1777006785-a5hsnq.jpg\"><\/p>\n\n\n\n<p>Options:<br>(1) Object A<br>(2) Object B<br>(3) Object C<br>(4) Object D \u2705<\/p>\n\n\n\n<p>Explanation: A net force acts on an object only when its velocity changes, i.e., when there is acceleration.<\/p>\n\n\n\n<p>In a position\u2013time graph:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Straight line \u2192 constant velocity \u2192 no net force<\/li>\n\n\n\n<li>Curved line \u2192 changing velocity \u2192 net force present<\/li>\n<\/ul>\n\n\n\n<p>From the given graph, the object with a curved position\u2013time graph is D.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.27: A sailor jumps out from a small boat to the shore. As the sailor jumps forward, will the boat move? If yes, in which direction and why.<\/strong><br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1777007940-sybgvg.jpg\"><\/p>\n\n\n\n<p>Solution: Yes, the boat will move.<\/p>\n\n\n\n<p>When the sailor jumps forward, he pushes the boat backward. According to Newton\u2019s Third Law of Motion, every action has an equal and opposite reaction.<\/p>\n\n\n\n<p>Thus, the forward force exerted by the sailor on the shore produces an equal and opposite force on the boat, causing it to move backward (opposite to the direction of the jump).<\/p>\n\n\n\n<p><strong>Conclusion:<\/strong> The boat moves backward due to action\u2013reaction forces.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.28: During a high jump event, a landing mat or sand bed is placed for the athlete to fall upon. Explain the reason behind it.<\/strong><br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1777008516-njv5vh.jpg\"><\/p>\n\n\n\n<p>Solution: A landing mat or sand bed increases the time of impact when the athlete falls.<br>According to the relation {tex}F=\\frac{\\Delta p}{\\Delta t}{\/tex}, if the time of contact {tex}\\Delta t{\/tex} increases, the force {tex}F{\/tex} decreases.<\/p>\n\n\n\n<p>Thus, the mat or sand reduces the force acting on the athlete&#8217;s body, preventing injury.<\/p>\n\n\n\n<p><strong>Conclusion:<\/strong> The landing mat\/sand bed reduces the impact force by increasing the time of collision.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.29: A hand cart loaded with vegetables collides with an identical but empty hand cart. During the collision:<\/strong><\/p>\n\n\n\n<p>Options:<br>(1) the loaded cart exerts a force of larger magnitude on the empty cart.<br>(2) the empty cart exerts a force of larger magnitude on the loaded cart.<br>(3) neither cart exerts a force on the other.<br>(4) the loaded cart and the empty cart, both exert an equal magnitude of force on each other. \u2705<\/p>\n\n\n\n<p>Explanation: According to Newton\u2019s Third Law of Motion, for every action, there is an equal and opposite reaction.<\/p>\n\n\n\n<p>During the collision, both carts exert forces on each other that are equal in magnitude and opposite in direction, regardless of their masses.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.30: The acceleration-mass graph for the acceleration produced by a force on objects of different masses is plotted in the figure. Plot the force-mass graph for this case.<\/strong><br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1777008873-que3fq.jpg\"><\/p>\n\n\n\n<p>Solution: From the given acceleration-mass graph, acceleration is inversely proportional to mass {tex}\\rightarrow a \\propto \\frac{1}{m}{\/tex}.<br>Thus, {tex}F=m a={\/tex} constant.<br>So, the force is constant for all masses.<br><strong>Force-mass graph:<\/strong><br>A horizontal straight line parallel to the mass axis.<br><strong>Conclusion: <\/strong>Force remains constant, independent of mass.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.31: The velocity-time graph of an object of mass 10 kg moving along a straight line is shown in the figure. Calculate the force acting on the object by using the graph.<\/strong><br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1777009066-ggwhx7.jpg\"><\/p>\n\n\n\n<p>Solution: From the velocity-time graph:<br>Acceleration {tex}a={\/tex} slope of v-t graph<\/p>\n\n\n\n<p>{tex} a=\\frac{v-u}{t}{\/tex}<\/p>\n\n\n\n<p>From the graph, velocity increases uniformly from {tex}0 {~m} \/ {s}{\/tex} to {tex}20 {~m} \/ {s}{\/tex} in 10 s<\/p>\n\n\n\n<p>{tex} a=\\frac{20-0}{10}=2 {~m} \/ {s}^2{\/tex}<\/p>\n\n\n\n<p>Given mass {tex}m=10 {~kg}{\/tex}<\/p>\n\n\n\n<p>{tex} F=m a=10 \\times 2=20 {~N}{\/tex}<\/p>\n\n\n\n<p>Force acting on the object {tex}=20 {~N}{\/tex}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.32: A bullet of mass 50 g moving with a speed of 100 m s<sup>-1<\/sup> enters a heavy stationary wooden block and stops after penetrating a distance of 50 cm. Estimate the stopping force acting on the bullet (assume that the bullet undergoes constant acceleration within the block).<\/strong><\/p>\n\n\n\n<p>Solution: Given:<br>Mass of bullet, {tex}m=50 g=0.05 {~kg}{\/tex}<br>Initial velocity, {tex}u=100\\ {m}\\ {s}^{-1}{\/tex}<br>Final velocity, {tex}v=0{\/tex}<br>Distance, {tex}s=50 \\ cm=0.5 m{\/tex}<br>Using equation of motion:<\/p>\n\n\n\n<p>{tex}v^2=u^2+2 a s{\/tex}<br>{tex}0=(100)^2+2 \\times a \\times 0.5{\/tex}<br>{tex}0=10000+a \\Rightarrow a=-10000 {~ms}^{-2}{\/tex}<\/p>\n\n\n\n<p>Now,<\/p>\n\n\n\n<p>{tex} F=m a=0.05 \\times(-10000)=-500 N{\/tex}<\/p>\n\n\n\n<p>Stopping force {tex}=500 {~N}{\/tex} (opposite to the direction of motion).<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.33: An ace footballer converted a penalty shot by kicking the football with a speed of 108 km h<sup>-1<\/sup>. The estimated force they imparted was 800 N. The mass of the football was 0.4 kg. Calculate the time of contact between their foot and the ball.<\/strong><\/p>\n\n\n\n<p>Solution: Given:<br>Speed {tex}v=108 {~km} \/ {h}=30 {~m} \/ {s}{\/tex}<br>Mass {tex}m=0.4 {~kg}{\/tex}<br>Force {tex}F=800 {~N}{\/tex}<br>Initial velocity {tex}u=0{\/tex}<br>Using {tex}F=\\frac{\\Delta p}{\\Delta t}=\\frac{m(v-u)}{t}{\/tex}<\/p>\n\n\n\n<p>{tex}t=\\frac{m(v-u)}{F}=\\frac{0.4 \\times 30}{800}{\/tex}<br>{tex}t=\\frac{12}{800}=0.015 {~s}{\/tex}<\/p>\n\n\n\n<p>Time of contact {tex}=0.015 {~s}{\/tex}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.34: An object of mass 2 kg moving with a constant velocity of 10 m s<sup>-1<\/sup> encounters a rough patch where the force of friction on the object is 7 N. At the same time, an additional constant force of 3 N opposing the motion is applied on the object. After entering the rough patch, how much distance does the object travel before coming to rest?<\/strong><\/p>\n\n\n\n<p>Solution: Given:<br>Mass {tex}m=2 {~kg}{\/tex}<br>Initial velocity {tex}u=10 {~m} \/ {s}{\/tex}<br>Final velocity {tex}v=0{\/tex}<br>Friction force {tex}=7 {~N}{\/tex}<br>Additional opposing force {tex}=3 {~N}{\/tex}<\/p>\n\n\n\n<p>Total opposing force {tex}F=7+3=10 {~N}{\/tex}<\/p>\n\n\n\n<p>Acceleration:<\/p>\n\n\n\n<p>{tex} a=\\frac{F}{m}=\\frac{-10}{2}=-5 {~m} \/ {s}^2{\/tex}<\/p>\n\n\n\n<p>Using:<\/p>\n\n\n\n<p>{tex}v^2=u^2+2 a s{\/tex}<br>{tex}0=(10)^2+2(-5) s{\/tex}<br>{tex}0=100-10 s \\Rightarrow s=10 {~m}{\/tex}<\/p>\n\n\n\n<p>Distance travelled before coming to rest = 10 m<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.35: A tractor pulls a harrow (a ploughing tool) of mass {tex}m_1{\/tex} with a net force {tex}F{\/tex} resulting in an acceleration of {tex}a_1{\/tex}. The same tractor pulls a trolley of mass {tex}m_2{\/tex} with a force {tex}F{\/tex} producing an acceleration of {tex}a_2{\/tex}. If the tractor now pulls the trolley with the harrow placed on it (with the same force {tex}F{\/tex}), then obtain an expression for the resulting acceleration in terms of {tex}a_1{\/tex} and {tex}a_2{\/tex}. Ignore friction.<\/strong><\/p>\n\n\n\n<p>Solution: Given:<br>Mass of harrow {tex}=m_1{\/tex}, acceleration {tex}=a_1{\/tex}<br>Mass of trolley {tex}=m_2{\/tex}, acceleration {tex}=a_2{\/tex}<br>Force applied in all cases {tex}=F{\/tex}<\/p>\n\n\n\n<p>For harrow:<\/p>\n\n\n\n<p>{tex} F=m_1 a_1 \\Rightarrow m_1=\\frac{F}{a_1}{\/tex}<\/p>\n\n\n\n<p>For trolley:<\/p>\n\n\n\n<p>{tex} F=m_2 a_2 \\Rightarrow m_2=\\frac{F}{a_2}{\/tex}<\/p>\n\n\n\n<p>When both are together:<br>Total mass {tex}=m_1+m_2{\/tex}<\/p>\n\n\n\n<p>{tex}=\\frac{F}{a_1}+\\frac{F}{a_2}{\/tex}<br>{tex}=F\\left(\\frac{1}{a_1}+\\frac{1}{a_2}\\right){\/tex}<\/p>\n\n\n\n<p>Let resulting acceleration {tex}=a{\/tex}<br>Using {tex}F=\\left(m_1+m_2\\right) a{\/tex}:<\/p>\n\n\n\n<p>{tex} F=F\\left(\\frac{1}{a_1}+\\frac{1}{a_2}\\right) a{\/tex}<\/p>\n\n\n\n<p>Cancel {tex}F{\/tex}:<\/p>\n\n\n\n<p>{tex}1=\\left(\\frac{1}{a_1}+\\frac{1}{a_2}\\right) a{\/tex}<br>{tex}a=\\frac{1}{\\frac{1}{a_1}+\\frac{1}{a_2}}{\/tex}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.36: When the pole of a bar magnet is brought close to a magnetic compass, the bar magnet and the compass needle (which is also a magnet) exert a magnetic force on each other. As per Newton\u2019s third law of motion, both the forces are equal in magnitude and opposite in direction. However, the compass needle moves, whereas the bar magnet does not move. Explain why.<\/strong><br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1777271884-y2nvyw.jpg\"><\/p>\n\n\n\n<p>Solution: When a bar magnet is brought near a compass needle, both exert equal and opposite forces on each other (Newton&#8217;s Third Law).<\/p>\n\n\n\n<p>However, motion depends on acceleration, not just force.<br>From Newton&#8217;s Second Law:<\/p>\n\n\n\n<p>{tex} a=\\frac{F}{m}{\/tex}<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The compass needle has very small mass, so even a small force produces large acceleration, hence it moves easily.<\/li>\n\n\n\n<li>The bar magnet has much larger mass (and is usually fixed or held), so the same force produces negligible acceleration, hence it does not move.<\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"<p>How Forces Affect Motion &#8211; NCERT Solutions Class 9 Science Exploration includes all the questions with solutions given in NCERT Class 9 Science Exploration textbook. NCERT Solutions Class 9 How Forces Affect Motion \u2013 NCERT Solutions Q.1: Why does a canoe move forward when the canoeist pushes water backwards with their paddle and why does &#8230; <a title=\"How Forces Affect Motion &#8211; NCERT Solutions Class 9 Science Exploration\" class=\"read-more\" href=\"https:\/\/mycbseguide.com\/blog\/how-forces-affect-motion-ncert-solutions-class-9-science-exploration\/\" aria-label=\"More on How Forces Affect Motion &#8211; NCERT Solutions Class 9 Science Exploration\">Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[281,2071,2072],"tags":[216],"class_list":["post-31534","post","type-post","status-publish","format-standard","hentry","category-ncert-solutions","category-ncert-solutions-class-9","category-ncert-solutions-class-9-science-exploration","tag-ncert-solutions"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.0 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>How Forces Affect Motion - NCERT Solutions Class 9 Science Exploration | myCBSEguide<\/title>\n<meta name=\"description\" content=\"How Forces Affect Motion - NCERT Solutions Class 9 Science Exploration includes all the questions with solutions given in Science Exploration\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mycbseguide.com\/blog\/how-forces-affect-motion-ncert-solutions-class-9-science-exploration\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"How Forces Affect Motion - NCERT Solutions Class 9 Science Exploration | myCBSEguide\" \/>\n<meta property=\"og:description\" content=\"How Forces Affect Motion - NCERT Solutions Class 9 Science Exploration includes all the questions with solutions given in Science Exploration\" \/>\n<meta property=\"og:url\" content=\"https:\/\/mycbseguide.com\/blog\/how-forces-affect-motion-ncert-solutions-class-9-science-exploration\/\" \/>\n<meta property=\"og:site_name\" content=\"myCBSEguide\" \/>\n<meta property=\"article:publisher\" content=\"https:\/\/www.facebook.com\/mycbseguide\/\" \/>\n<meta property=\"article:published_time\" content=\"2026-05-20T11:40:11+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2026-05-20T11:40:52+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1776947018-f3v2vx.jpg\" \/>\n<meta name=\"author\" content=\"myCBSEguide\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:creator\" content=\"@mycbseguide\" \/>\n<meta name=\"twitter:site\" content=\"@mycbseguide\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"myCBSEguide\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"22 minutes\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"Article\",\"@id\":\"https:\/\/mycbseguide.com\/blog\/how-forces-affect-motion-ncert-solutions-class-9-science-exploration\/#article\",\"isPartOf\":{\"@id\":\"https:\/\/mycbseguide.com\/blog\/how-forces-affect-motion-ncert-solutions-class-9-science-exploration\/\"},\"author\":{\"name\":\"myCBSEguide\",\"@id\":\"https:\/\/mycbseguide.com\/blog\/#\/schema\/person\/10b8c7820ff29025ab8524da7c025f65\"},\"headline\":\"How Forces Affect Motion &#8211; 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