Advertisements
Advertisements
प्रश्न
A woman throws an object of mass 500 g with a speed of 25 ms1.
- What is the impulse imparted to the object?
- If the object hits a wall and rebounds with half the original speed, what is the change in momentum of the object?
उत्तर
Mass of the object (m) = 500 g = 0.5 kg
Speed of the object (v) = 25 m/s
a. Impulse imparted to the object = Change in momentum
= mv – mu
= m(v – u)
= 0.5(25 – 0)
= 12.5 N-s
b. Velocity of the object after rebounds
= `- 25/2` m/s
v' = – 12.5 m/s
∴ Change in momentum = m(v’ – v)
= 0.5 (– 12.5 – 25)
= – 18.75 N-s
APPEARS IN
संबंधित प्रश्न
A man of mass 70 kg stands on a weighing scale in a lift which is moving
- upwards with a uniform speed of 10 m s-1
- downwards with a uniform acceleration of 5 m s–2
- upwards with a uniform acceleration of 5 m s–2. What would be the readings on the scale in each case?
- What would be the reading if the lift mechanism failed and it hurtled down freely under gravity?
The below figure shows the position-time graph of a particle of mass 4 kg.
- What is the force on the particle for t < 0, t > 4 s, 0 < t < 4 s?
- What is the impulse at t = 0 and t = 4 s? (Consider one-dimensional motion only.)
A block of mass m is placed on a smooth wedge of inclination θ. The whole system is accelerated horizontally so that the block does not slip on the wedge. The force exerted by the wedge on the block has a magnitude.
A person says that he measured the acceleration of a particle to be non-zero even though no force was acting on the particle.
Find the acceleration of the block of mass M in the situation shown in the following figure. All the surfaces are frictionless and the pulleys and the string are light.
A monkey of mass 15 kg is climbing a rope fixed to a ceiling. If it wishes to go up with an acceleration of 1 m/s2, how much force should it apply on the rope? If the rope is 5 m long and the monkey starts from rest, how much time will it take to reach the ceiling?
The correct form of Newton's second law is :
A ball is thrown vertically upwards. It returns 6 s later. Calculate : The greatest height reached by the ball . (Take g = 10 m s-2)
An electron of mass 9 × 10−31 kg is moving with a linear velocity of 6 × 107 ms−1. Calculate the linear momentum of electron.
The INCORRECT statement about Newton's second law of motion is
