CBSE PHYSICS: Enrich your study: IX:Force and laws of motion sol...: 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...
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
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.
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.
More related popular posts
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.
More related popular posts
9th Forces and Laws of motion teat paper-1 Download File
9th force and law of motion Test paper -2 Download File
9th Test papers unsolved-3 Download File
9th Numerical unsolved Download File
Solved Numerical Problems Download File
Very useful information shared and students can prepare easily for examination by this questionnaire.Online tutoring is really helpful in preparing for exams.
ReplyDeleteMomentum Equation