CBSE PHYSICS: IX:Force and laws of motion solved Problems

1. An athlete runs a certain rest before taking a long jump. Why?

Solution: An athlete runs a certain distance to accelerate himself and gain enough momentum so that he can jump through the maximum possible length.

2. Springs are provided in car seats. Why?

Solution: The springs in the car seats absorb shock (sudden jumps) due to the roughness of the road. Thus, making the ride more comfortable.

3. A gun of mass 1500 kg fires a shell of mass 15kg with velocity 150 m/s. calculate velocity of recoil of the gun.

Solution: Before firing, total momentum of the gun and shell is = 0 (they were all at rest)

After firing, the momentum of the shell = 15 X 150 = 2250 Ns

The momentum of the gun is = 1500v Ns

By conservation of momentum,

Total momentum before firing = Total momentum after firing

0 = 2250 + 1500v

v = -2250/1500 = -1.5 m/s

4. Why cricketer pulls his hands backwards while catching the ball

Solution: cricketer pulls his hands backwards while catching the ball. When he does so, momentum of the ball reduces slowly, and time t required for this increases. As per F =DP/t , as t increase, magnitude of F decreases. As a result cricketer can catch the ball, easily, without any injury.

5. Why A karate player in order to break a brick, hits it quickly

Solution: A karate player in order to break a brick, hitsit quickly, so in a short time there is a large change

in momentum and as per F =change in p/t as large force acts on the brick and it breaks.in momentum and as per F =change in p/t as large force acts on the brick and it breaks.in momentum and as per F =change in p/t as large force acts on the brick and it breaks.

Solution: A karate player in order to break a brick, hitsit quickly, so in a short time there is a large change

in momentum and as per F =change in p/t as large force acts on the brick and it breaks.in momentum and as per F =change in p/t as large force acts on the brick and it breaks.in momentum and as per F =change in p/t as large force acts on the brick and it breaks.

6. What is impulse of Force?

I = F x t= change in p

Like force, impulse of force is also a vector quantity. Direction of impulse of force is same as the direction of force. Unit of impulse of force is kg m/s or Ns.Like force, impulse of force is also a vector quantity. Direction of impulse of force is same as the direction of force. Unit of impulse of force is kg m/s or Ns.

7. A ball of 150 g is thrown at 20 m/s towards the batsman. He hits the ball in the direction oppsite to intial direction of motion with velocity 25 m/s. If the ball is hit in 0.01 s, calculate change in momentum of the ball and force applied by the batsman on the ball.

Solution: m = 150 g = 150/1000kg= 0.15 kg

u = 20 m/s, v = 25 m/s, t = 0.01 s

Initial momentum of the ball

pi = mu = (0.15)(20) = 3 kgm/s

Final momentum of the ball after being hit

pf = mv = (0.15)( - 25) = - 3.75 kgm/s

Here negative sign indicates that direction of motion of the ball after hitting is opposite to initial

direction of motion of the balls.

Change in momentum of the ball Dp = pf - pi = ( - 3.75 - 3) = - 6.75 kgm/s

So change in momentum of the bat = 6.75 kgm/s

Force applied by the bat,

F =Change in p/t= 6.75/ 0.01 = 675 kg m/s2 = 675 N

8.A bullet of 20 g is fired horizontally from a pistol of 2 kg mass with velocity 150 ms-1. How much would be the velocity of pistol in backward direction after firing the bullet ?

Solution : Mass of bullet m1 = 20 g = 0.02 kg

Mass of pistol m2 = 2 kg

Initial velocity of bullet u1 = 0 m/s

Initial velocity of pistol u2 = 0 m/s

Final velocity of the bullet = v1 = 150 ms-1

Final velocity of the pistol v2 = ?

According to law of conservation of momentum

m1v1 + m2v2 = m1u1 + m2u2

(0.02)(150) + (2)(v2) = (0.02)(0) + (2)(0)

v2 = - (0.02)(150)/2 = - 1.5 ms–1

Here, negative sign indicates that motions ofpistol and bullet are in opposite direction.

9th Forces and Laws of motion teat paper-1 Download File

Solution: m = 150 g = 150/1000kg= 0.15 kg

u = 20 m/s, v = 25 m/s, t = 0.01 s

Initial momentum of the ball

pi = mu = (0.15)(20) = 3 kgm/s

Final momentum of the ball after being hit

pf = mv = (0.15)( - 25) = - 3.75 kgm/s

Here negative sign indicates that direction of motion of the ball after hitting is opposite to initial

direction of motion of the balls.

Change in momentum of the ball Dp = pf - pi = ( - 3.75 - 3) = - 6.75 kgm/s

So change in momentum of the bat = 6.75 kgm/s

Force applied by the bat,

F =Change in p/t= 6.75/ 0.01 = 675 kg m/s2 = 675 N

8.A bullet of 20 g is fired horizontally from a pistol of 2 kg mass with velocity 150 ms-1. How much would be the velocity of pistol in backward direction after firing the bullet ?

Solution : Mass of bullet m1 = 20 g = 0.02 kg

Mass of pistol m2 = 2 kg

Initial velocity of bullet u1 = 0 m/s

Initial velocity of pistol u2 = 0 m/s

Final velocity of the bullet = v1 = 150 ms-1

Final velocity of the pistol v2 = ?

According to law of conservation of momentum

m1v1 + m2v2 = m1u1 + m2u2

(0.02)(150) + (2)(v2) = (0.02)(0) + (2)(0)

v2 = - (0.02)(150)/2 = - 1.5 ms–1

Here, negative sign indicates that motions ofpistol and bullet are in opposite direction.

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Very useful information shared and students can prepare easily for examination by this questionnaire.Online tutoring is really helpful in preparing for exams.

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