Question

A magnet of mass $2\text{kg}$ is thrown with a velocity of $5\text{m/s}$ towards a moving metal block of unknown mass. Both magnet and block are moving along a smooth horizontal surface. After the collision, the magnet sticks to the metal block. If the initial momentum of system is $48\text{kg.m/s}$ and the initial velocity of $\text{COM}$ of system is $3\text{m/s}$, then what is the magnitude of initial velocity of the metal block?

• A. $2.7\text{m/s}$
• B. $3.92\text{m/s}$
• C. $1.92\text{m/s}$
• D. $4.92\text{m/s}$

Answer 1

The correct option is A $2.7\text{m/s}$

Let $v=$ Initial velocity of block
$m=$ mass of block
$m_2=$ $2\text{kg}=$mass of magnet
$v_2=5\text{m/s}=$ Initial velocity of magnet
Initial velocity of $\text{COM}$ of system $v_{\text{CM}}=3\text{m/s}$

Total mass of system$=(m+2)\text{kg}$
Initial momentum of system:
$\Rightarrow P_i=(m+2)\times v_{\text{CM}}$

Or, $48=3m+6$
$\therefore m=\frac{42}{3}=14\text{kg}$
Therefore mass of the metal block is $m_1=14\text{kg}$.

Applying formula for velocity of $\text{COM}$ of system :
$v_{\text{CM}}=\frac{m_1{v}_1+m_2{v}_2}{m_1+m_2}$

$\Rightarrow 3=\frac{14v+(2\times 5)}{14+2}$
$\Rightarrow 48=14v+10$

$\therefore v=\frac{38}{14}=2.7\text{m/s}$

Hence magnitude of initial velocity of the metal block is $2.7\text{m/s}$.