Question

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

• A. $10{kgms}^{-1}$​,$10{kgms}^{-1}$​
• B. $20{kgms}^{-1}$​,$10{kgms}^{-1}$​
• C. $5{kgms}^{-1}$​,$10{kgms}^{-1}$​
• D. $10{kgms}^{-1}$​,$5{kgms}^{-1}$

Answer 1

The correct option is A $10{kgms}^{-1}$​,$10{kgms}^{-1}$​

Let, mass of the block be $m_1$ and and velocity of the block be $u_1$
Let, mass of the stationary wooden block be $m_2$ and velocity of the stationary block be $u_2$
Given, $m_1=1kg$, $u_1=10m{s}^{-1}$ and $m_2=5kg$, $u_2=0$(since at rest)
The momentum before collision = $m_1{u}_1+m_2{u}_2=10\times 1+0=10{kgms}^{-1}$​

From the law of conservation of momentum, momentum before collision should be equal to momentum after the collision as no external force was acting on the bodies.
$\therefore$The momentum after collision = $10{kgms}^{-1}$​