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Last Updated On [03-11-2023]

In the arrangement shown in Fig., the ball and the block have the same mass  1 kg each,  and length  m. Coefficient of friction between the block and the floor is 0.5. When the ball is released from the position shown in Fig., it collides with the block and the block stops after moving a distance 2.50 m

The velocity of block just after collision is

a)

10 m/s

b)

5 m/s

c)

2.5 m/s

d)

3 m/s


Question ID - 101257 :-

In the arrangement shown in Fig., the ball and the block have the same mass  1 kg each,  and length  m. Coefficient of friction between the block and the floor is 0.5. When the ball is released from the position shown in Fig., it collides with the block and the block stops after moving a distance 2.50 m

The velocity of block just after collision is

a)

10 m/s

b)

5 m/s

c)

2.5 m/s

d)

3 m/s

1 Answer
5876 Votes
3537

Answer Key : (b) -

(b)

When ball is released, it moves along a vertical circle with centre at . Kinetic energy of ball just before collision is equal to loss of its potential energy from point of release to the point of collision. Therefore, velocity  of ball, just before collision, is given by

After collision, the block starts to move towards right. But it is retarded by force of friction and ultimately it comes to rest. According to law of conservation of energy,

Kinetic energy of the block just after collision = Work done by it against friction

Therefore, its velocity just after collision is given by

Where and

Coefficient of restitution,

Where and

or   (i)

Applying law of conservation of momentum,

   or

Substituting in Eq. (i), we get



Next Question :

First two terms are connected by some relationship.  The same relationship is applicable for the next term in which one is blank space.  Identify the suitable term.

Earth : Moon :: Sun : __________

(a)

Star

(b)

Earth

(c)

Sky

(d)

Moon


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