9th Physics Ch 8 Force and Laws of Motion PDF

Grade - 9th Chapter No. – 8 Subject: Physics Chapter Name – Force and Laws of Motion Numericals for Practice Part – 1 Numerical Examples...

Grade - 9th Chapter No. – 8 Subject: Physics Chapter Name – Force and Laws of Motion Numericals for Practice Part – 1 Numerical Examples based on Momentum and Newton's Second Law of Motion Q.1 What is the momentum of a man of mass 75 kg, when he walks with a uniform velocity of 2 m/s? Ans. Here, m = 75 kg, v = 2 m/s Momentum, p = mv = 75 kg x 2 m/s = 150 kg m/s. Q.2 By how much does the momentum of a body of mass 5 kg change when its speed (i) decreases from 20 m/s to 0.20 m/s; and (ii) increases from 30 m/s to 40 m/s? Ans. (i) m = 5 kg, u = 20 m/s, v = 0.20 m/s Change in momentum = m (v - u) = 5(0.20 - 20) kg m/s = - 99 kg m/s The negative sign shows that the momentum decreases. (ii) m = 5 kg, u = 30 m/s, v = 40 m/s Change in momentum = m (v - u) = 5(40 - 30) = 50 kg m/s Here, the momentum of the body increases. Q.3 A man throws a ball of mass 0.5 kg vertically upwards with a velocity of 10 m/s. What will be its initial momentum? What would be its momentum at the highest point of its reach? Ans. m = 0.5 kg, u = 10 m/s Initial momentum = mu = 0.5 x 10 = 5 kg m/s At the highest point, velocity of ball is zero, so momentum of the ball = 0.5 × 0 = 0 Q.4 A steam engine of mass 3 x 104 kg pulls two wagons each of mass 2 × 104 kg with an acceleration of 0.2 m/s2. Neglecting frictional force, calculate the: (i) force exerted by the engine. (ii) force experienced by each wagon. Ans. Total mass, m = 3 × 104 + 2 x 2 × 102 = 7 x 104 kg, a = 0.2 m/s2 (i) F = ma = 7 x 104 x 0.2 = 1.4 x 104 N (i) Force experienced by each wagon = 1.4 x 104 N Q.5 A man pushes a box of mass 50 kg with a force of 80 N. What will be an acceleration of the box due to this force? What would be an acceleration if the mass is halved? Ans. In first case: m = 50 kg, F = 80 N 𝐹 80 𝑚 Acceleration, a = = = 1.6 𝑚 50 𝑠2 In second case: m =25 kg, F = 80 N 𝐹 80 𝑚 Acceleration, a = = = 3.2 𝑚 25 𝑠2 Q.6 Calculate the force required to impart to a car a velocity of 30 m/s in 10 s. The mass of the car is 1500 kg. Ans. Here, u = 0, v = 30 m/s, t = 10 s 𝑣−𝑢 30−0 30 𝑚 Acceleration, 𝑎 = = = =3 2 𝑡 10 10 𝑠 Now m = 1500 kg, a = 3 m/s2 Required force, F = ma = 1500 x 3N = 4500 N. Q.7 A cricket ball of mass 70g moving with a velocity of 0.5 m/s is stopped by a player in 0.5 s. What is the force applied by the player to stop the ball? Ans. Here, m = 70 g = 0.070 kg, u = 0.5 m/s, v = 0, t = 0.5 s 𝑣−𝑢 0−0.5 Force, F = ma = 𝑚 ( ) = 0.070 ( ) = −0.070 𝑁 𝑡 0.5 Negative sign shows the retarding nature of the force. Q.8 A car of mass 1000 kg moving with a velocity of 45 km/h collides with a tree and comes to a stop in 5 s. What will be the force exerted by the car on the tree? 45 𝑋 1000 25 Ans. Here, m = 1000 kg, u = 45 km/h = = 𝑚/𝑠 v = 0, t = 5 s 3600 2 25 𝑣−𝑢 0− 2 Force, F = ma = 𝑚 ( ) = 1000 ( ) = −2500 𝑁 𝑡 5 Negative sign shows the retarding nature of the force. Q.9 A force of 10 N accelerates an object from 15 m/s to 25 m/s in 20 s. Calculate the mass of the object. Ans. Here F = 10 N, u = 15 m/s, v = 25 m/s, t = 20 s 𝑣−𝑢 25 − 15 10 1 𝑚 𝑚 Acceleration, 𝑎 = = = = 2 = 0.5 𝑡 20 20 2 𝑠 𝑠2 𝐹 10 Mass, m = = 1 = 20 𝑘𝑔 𝑎 ⁄2 Q.10 For how much time should the force of 400 N be exerted on a body of mass 8 kg to increase its velocity from 150 m/s to 300 m/s? Ans. F = 400 N, m = 8 kg, u = 150 m/s, v = 300 m/s, t = ?, a = ? 𝐹 400 𝑚 Acceleration, a = = = 50 2 𝑚 8 𝑠 𝑣−𝑢 300−150 150 Time, t = = = =3𝑠 𝑎 50 50 Q.11 A body of mass 4 kg is moving with a uniform velocity of 5 m/s. It is acted upon by a force of 12 N. What will be its velocity after 1 s? Ans. m = 4 kg, F = 12 N 𝐹 12 𝑚 Acceleration, a = = = 3 2 𝑚 4 𝑠 Now, u = 5 m/s, a = 3 m/s2, t = 1 s Final velocity, v = u + at = 5 + (3 x 1) = 8 m/s. Q.12 Two blocks made of different metals identical in shape and size are acted upon by equal forces which cause them to slide on a horizontal surface. The acceleration of the second block is found to be 5 times that of the first, what is the ratio of the mass of the second to the first? Ans. Let the masses of the two blocks be m1, and m2, respectively, and a and 5a be their respective accelerations. For first block 𝐹1 = 𝑚1. 𝑎 For second block 𝐹2 = 𝑚2. 5𝑎 Now, equal forces are acting upon on it, F1 = F2 𝑚1. 𝑎 = 𝑚2. 5𝑎 𝑚1 5 = =5∶1 𝑚2 1 Q.13 A driver accelerates his car first at the rate of 1.8 m/s2 and then at the rate of 1.2 m/s2. Calculate the ratio of the force exerted by the engine in the two cases. 𝐹1 𝑚𝑎1 1.8 18 3 Ans. Let m be the mass of the car. Then = = = = 𝐹2 𝑚𝑎2 1.2 12 2 Therefore, the ratio is 3 : 2