Home
Class 11
PHYSICS
The 4 kg sphere form rest when theta = 6...

The `4 kg` sphere form rest when `theta = 60^(@)` strikes a block mass of `5 kg` placed on a rough horizontal plane and comes to rest after collision. The `5 kg` block comes to rest after moving a distance of `0.8m`. Find `mu` of ground & block & coefficient of restituation e.

Text Solution

Verified by Experts

The correct Answer is:
`e = 0.8, mu = 0.4`
`u_(s) = sqrt(2g(1 - cos 60^(@))) = sqrt(2g xx (1)/(2)) = 3.13 m//s`
`COLM rArr 5u_(B) = 4 xx u_(s) rArr u_(B) = (4)/(5) sqrtg = 2.53 m//s`
Energy equation `(1)/(2)m ((4)/(5) sqrtg)^(2) = mu mg xx 0.8 rArr mu = 0.4`
`e = -((v_(2) - v_(1))/(u_(2) - u_(1))) = 0.8`
Promotional Banner

Topper's Solved these Questions

  • CENTRE OF MASS

    ALLEN |Exercise EXERCISE-V A|20 Videos

  • CENTRE OF MASS

    ALLEN |Exercise EXERCISE-V B|19 Videos

  • CENTRE OF MASS

    ALLEN |Exercise EXERCISE-IV A|32 Videos

  • BASIC MATHEMATICS USED IN PHYSICS &VECTORS

    ALLEN |Exercise EXERCISE-IV ASSERTION & REASON|11 Videos

  • ELASTICITY, SURFACE TENSION AND FLUID MECHANICS

    ALLEN |Exercise Exercise 5 B (Integer Type Questions)|3 Videos

Similar Questions

Explore conceptually related problems

A force of 50N acts in the direction as shown in figure. The block of mass 5kg, resting on a smooth horizontal surface. Find out the acceleration of the block.

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

A block is released from rest from point on a rough inclined place of inclination 37^(@). The coefficient of friction is 0.5.

A block is kept on a rough horizontal surface as shown. Its mass is 2 kg and coefficient of friction between block and surface (mu)=0.5. A horizontal force F is acting on the block. When

A block of mass 3kg is at rest on a rough inclined plan as shown in the Figure. The magnitude of net force exerted by the surface on the block will be

A block of mass m=1/3 kg is kept on a rough horizontal plane. Friction coefficient is mu=0.75 . The work done by minimum force required to drag to the block along the plane by a distance 5m is :-

A block of mass 2 kg is kept on a rough horizontal floor an pulled with a force F. If the coefficient of friction is 0.5. then the minimum force required to move the block is :-

A block of mass 15 kg is resting on a rough inclined plane as shown in figure ,The block is tied by a horizontal string which has a tension 50 N The coefficient of friction of contact is

A bullet of mass 50g moving with velocity 600m//s hits block of mass 1.0kg placed on a rough horizontal ground and comes out of the block with a velocity of 400m//s . The coefficient of friction between the block and the ground is 0.25 . Neglect loss of mass of the block as the bullet pierces through it. (a) In spite of the fact that friction acts a an external force, can you apply principle of conservation of momentum during interaction of the ullet with the block? (b) Find velocity of the block immediately after the bullet pierces through it. (c)FInd the distance the block will travel before it stops.