Question

A block mass $M=5kg$ is moving on a horizontal plane and an object of mass $m=1kg$ is dropped on to the block hitting it with a vertical velocity of $v_1=10m/s$. If the speed of the block at the same instant is $v_2=2m/s$ on the floor and collision is momentary and the object sticks to the block on collision. Then what will be the speed of the block just after the collision if the coefficient of friction between the block and the horizontal plane is 0.4 ?

• A. $0.8m/s$
• B. $1.0m/s$
• C. $1.4m/s$
• D. $1.67m/s$

Answer 1

The correct option is B $1.0m/s$

Applying the impulse momentum theorem :
(a) In the vertical direction for mass $m$ : $\Rightarrow \int Ndt=0-(-m{v}_1)=10.........\left(i\right)$ (taking downward direction as positive)

(b) In the horizontal direction for mass $m$ :
$\Rightarrow -\mu \int Ndt=(M+m)v-\left(M{v}_2\right)....\left(ii\right)$ where $v$ is the final velocity of the system.
$\Rightarrow -0.4\times 10=\left(6\right)v-(5\times 2)....\left(2\right)$
On solving, $v=1.0m/s$