Question

Two blocks of equal masses are placed on a horizontal surface. The surface of A is smooth but that of B has a friction coefficient of 0.2 with the floor. Block A is given a speed 5 m/s, towards B which is kept at rest. Find the distance travelled by B, if the collision is perfectly elastic.
(take g = 10 $m/s^2$ )

• A. $\frac{5}{4}$ m
• B. $\frac{5}{4}$ m
• C. $\frac{5}{4}$ m
• D. $\frac{5}{2}$ m

Answer 1

The correct option is C $\frac{5}{4}$ m


Given
Initial speed of Block A = u = 5 m/s
Just after the collision speed of Block B is
By momentum conservation along the motion of the blocks (because no external force is present along that direction)
$mu+0=m{v}_1+m{v}_2$

After collision

Assume
$v_1=$ Speed of Block A towards right
$v_2=$ Speed of Block B towards right
Hence, $v=v_1+v_2$ ...(I)
as we know, $e=\frac{\text{speed of seperation}}{\text{speed of approact}}$
$e=\frac{v_2-v_1}{u-0}=1$ (because elastic collision e = 1)
$v_2-v_1=u$ ...(II)
by (I) and (II)
$v_2=u,v_1=0$
a friction force is acting on Block B,
$v^2-{\mu }^2=2as$
Here final speed v = 0
$\mu =5$ m/s
$a=\frac{\mu mg}{m}=-\mu g=-0.2\times 10$
$=-2m/s^2$
$0-5^2=2\times (-2)\times S$
force is acting in opposite direction of motion.
$S=\frac{25}{4}m.$