Question

A block of mass $M=5\text{kg}$ is moving on a horizontal plane. An object of mass $m=1\text{kg}$ is dropped on to the block hitting it with a vertical velocity of $v_1=10\text{m/s}.$ The speed of the block at the same instant is $v_2=2\text{m/s}$ along the floor. The collision is momentary and the object sticks to the block upon collision. What will be the speed of the block just after the collision if the co-efficient of friction between the block and the horizontal plane is $\mathrm{0.4.}$

• A. $1.4\text{m/s}$
• B. $0.8\text{m/s}$
• C. $1\text{m/s}$
• D. $1.67\text{m/s}$

Answer 1

The correct option is C $1\text{m/s}$

Applying the Impulse-momentum theorem:

In the vertical direction for $m$:

$\int Ndt=0-(-m{v}_1).....\left(1\right)$

(taking downward direction as positive)

In the horizontal direction:

$-\mu \int Ndt=(M+m)v-\left(M{v}_2\right)....\left(2\right)$

where, $v$ is the final velocity of the system.

Solving equations $\left(1\right)$ and $\left(2\right)$ we have,

$v=1\text{m/s}$

Hence, option $\left(B\right)$ is the correct answer.