Home
Class 12
PHYSICS
There are two blocks A and B placed on a...

There are two blocks A and B placed on a smooth surface. Block A has mass 10 kg and its is moving with velocity 0.8 m/s towards stationary B of unknown mass. At the time of collision, their velocities are given by the following graph :

Coefficient of restitution of the collision is

A

1.5

B

1

C

0.5

D

0.8

Text Solution

Verified by Experts

The correct Answer is:
B

`m_(A)xx0.8 =m_(A)xx0.2 +m_(0)xx1.0`
`m_(A)xx0.6 m_(B)xx1.0 m_(B)=0.6 m_(A)`
`e=(1-0.2)/(0.8)=1=1.5`
`I_(d)=6xx0.5-6xx0=3N-5=10xx(0.8-0.5)=10xx0.3 = 3NS`
`Delta U=(1)/(2)xx10xx(0.8)^(2)-(1)/(2)xx16xx(0.5)^(2)=5`
`xx0.64 8 xx0.25=3.2 -2.0=1.2J`.
Promotional Banner

Topper's Solved these Questions

  • DAILY PRACTICE PROBLEM

    RESONANCE ENGLISH|Exercise DPP No.66|20 Videos
  • DAILY PRACTICE PROBLEM

    RESONANCE ENGLISH|Exercise DPP No.67|10 Videos
  • DAILY PRACTICE PROBLEM

    RESONANCE ENGLISH|Exercise DPP No.64|20 Videos
  • CURRENT ELECTRICITY

    RESONANCE ENGLISH|Exercise High Level Problems (HIP)|19 Videos
  • ELECTRO MAGNETIC WAVES

    RESONANCE ENGLISH|Exercise Exercise 3|27 Videos

Similar Questions

Explore conceptually related problems

There are two blocks A and B placed on a smooth surface. Block A has mass 10 kg and its is moving with velocity 0.8 m/s towards stationary B of unknown mass. At the time of collision, their velocities are given by the following graph : Impulse of deformation is :

There are two blocks A and B placed on a smooth surface. Block A has mass 10 kg and its is moving with velocity 0.8 m/s towards stationary B of unknown mass. At the time of collision, their velocities are given by the following graph : Maximum deformation potential energy is :

A ball weighing 4 kg and moving with velocity of 8 m/sec. Collides with a stationary ball of mass 12 kg. After the collision, the two balls move together. The final velocity is

A sphere of mass m moving with a constant velocity u hits another stationary sphere of the same mass. If e is the coefficient of restitution, then ratio of velocities of the two spheres after collision will be

A block having mass m collides with an another stationary block having mass 2m. The lighter block comes to rest after collision. If the velocity of first block is v, then the value is coefficient of restitution will must be

Two point prticles A and B are placed in line on frictionless horizontal plane. If particle A (mass 1 kg ) is move with velocity 10 m//s towards stationary particle B (mass 2kg ) and after collision the two move at an angle of 45^(@) with the initial direction of motion, then find a. velocites of A and B just after collision. b. coefficient of restitution

A moving block having mass m , collides with another stationary block having mass 4m . The lighter block comes to rest after collision. When the initial velocity of the block is v , then the value of coefficient of restitution ( e) will be

Sphere A of mass m moving with a constant velocity u hits another stationary sphere B of the same mass. If e is the co-efficient of restitution, then ratio of velocities of the two spheres v_(A):v_(B) after collision will be :

A body of mass 10 kg and velocity 10 m/s collides with a stationary body of mass 5 kg . After collision both bodies stick to each other , velocity of the bodies after collision will be

Two masses 'm' and '2m' are placed in fixed horizontal circular smooth hollow tube as shown. The mass 'm' is moving with speed 'u' and the amss '2m' is stationary . After their first collision, the time elapsed for next collision. ( coefficient of restitution e=1//2)