*m*) on a

*smooth*horizontal surface. B and C which are initially at rest, are connectd by a spring of negligible mass and force constant

*K*. The block A is initially moving with velocity ‘v’ along the line joining B and C and hence it collides with B

*elastically*.

Now, answer the following questions related to the system:

**(1) After the collision the velocity of the block A is **

**(a) v/3**

**(b) ****–**** v**

**(c) ****–**** v/3**

**(d) v/2**

**(e) zero **

At the moment the block A collides with the block B, you need consider these two blocks only. In one dimensional *elastic* collision between two bodies of the *same* mass, if one body is initially at rest, the moving body comes to rest and the body initially at rest moves with the velocity of the moving body. After the collision, the velocity of A is therefore zero.

[Generally, if the moving block 1 has mass *m*_{1} and it moves with initial velocity v_{1i} towards_{ }the *stationary *block 2 of mass *m*_{2}, the final velocities of blocks 1 and 2 are given by

v_{1f} = v_{1i}(*m*_{1} – *m*_{2}) /(*m*_{1} + *m*_{2}) and

v_{2f} = 2 *m*_{1}v_{1i }/(*m*_{1} + *m*_{2})

The above relations can be easily obtained from momentum and kinetic energy conservation equations.

**(2) When the compression of the spring is maximum, the velocity of block B is**

**(a) v**

**(b) ****–**** v**

**(c) ****–**** v/3**

**(d) v/2**

**(e) zero **

On receiving the entire momentum (*m*v) of the block A, the block B moves towards the block C, compressing the spring. When the compression of the spring is maximum, both B and C move with the *same* velocity (v’). Since B and C have the *same *mass (*m*) and momentum has to be conserved, we have

*m*v = (*m* + *m*) v’

Therefore v’ = v/2.

**(3) When the compression of the spring is maximum, the kinetic energy of the system is **

**(a) (½) mv^{2}**

**(b) (1/4) mv^{2}**

**(c) ****Zero**

**(d) mv^{2}**

**(e) (1/8) mv^{2}**

The block A has come to rest after colliding with block B. When the compression of the spring is maximum, both B and C are moving with the *same* velocity v/2 as shown above. Therefore, the kinetic energy of the system under this condition is 2×(1/2)*m*(v/2)^{2} = **(1/4) mv^{2}**. [Note that the spring has negligible kinetic energy since its mass is negligible].

**(4) The maximum compression of the spring is **

**(a) √( mv^{2}/2K)^{}**

**(b) √( mv/2K)**

**(c) mv/2K**

**(d) mv^{2}/2K **

**(e) √(2 mv^{2}/K) **

The total energy of the system is equal to the initial kinetic energy (½ )*m*v^{2} of the colliding block A. Therfore we have

(½)*m*v^{2} = KE of A + KE of B + KE of C + PE of spring. The KE of A is zero after the collision. As shown above, the total KE of B and C is (1/4) *m*v^{2} when the compression of the spring is maximum. Therfore, the potential energy (PE) of the spring at maximum compression is given by the equation,

(½)*m*v^{2} = (1/4) *m*v^{2} + (1/2)*Kx*^{2} where *x *is the *maximum* compression of the spring.

This gives (1/2)*Kx*^{2} = (1/4)* m*v^{2} from which *x* = √(*m*v^{2}/2*K*).

You will find similar questions (with solution) from different branches of physics at **apphysicsresources.blogspot.com **

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