Question

A block of mass $1kg$ travelling in a straight line with a velocity of $10m{s}^{-1}$ collides and sticks to a stationary wooden block of mass $5kg$. After this, they both move together in the same straight line. Calculate the total momentum just before the impact, just after the impact and the velocity of the combined object.

• A. 10 kg ms^-1, 20 kgms^-1, 1.67 ms^-1
• B. 10 kg ms^-1, 10 kg ms^-1, 2 ms^-1
• C. 10 kg ms^-1, 10 kg ms^-1, 1.67 ms^-1
• D. 5 kg ms^-1, 10 kg ms^-1, 2 ms^-1

Answer 1

The correct option is C

10 kg ms^-1, 10 kg ms^-1, 1.67 ms^-1

Let $m_1$ is mass of first block, $m_2$ is mass of wooden block, $u_1$ is initial velocity of first block, $u_2$ is initial velocity of wooden block.
$m_1=1kg,u_1=10m{s}^{-1},m_2=5kg,u_2=0m{s}^{-1}$
The momentum before collision = m₁u₁+m₂u₂ = $10kg.m{s}^{-1}$.

The momentum after collision = (m₁+m₂)$\times v=6v$.

From the law of conservation of momentum, momentum before the collision should be equal to the momentum after the collision as no external force is acting on the bodies.
So, $6v=10$
$\Rightarrow v=1.67m{s}^{-1}$