A body of mass 'm' moving with speed $3 \mathrm{~m} / \mathrm{s}$ collides with a body of mass ${ }^{4} 2…
A body of mass 'm' moving with speed $3 \mathrm{~m} / \mathrm{s}$ collides with a body of mass ${ }^{4} 2 \mathrm{~m}^{\prime}$ at rest. The coalesced mass will start to move with a speed of
$3 \mathrm{~m} / \mathrm{s}$
$6 \mathrm{~m} / \mathrm{s}$
$9 \mathrm{~m} / \mathrm{s}$
$1 \mathrm{~m} / \mathrm{s}$
Solution
Initial velocity of moving mass $\mathrm{u}_{1}=3 \mathrm{~km} / \mathrm{h}$
Initial velocity of stationary mass $\mathrm{u}_{2}=0$
Thus momentum of the system initially $P_{i}=m_{1} u_{1}+m_{2} u_{2}$
$\therefore P_{i}=3 m+0=3 m$
Let the velocity of coalesced mass be V.
Thus momentum of the system finally $\mathrm{P}_{\mathrm{f}}=\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right) \mathrm{V}=3 \mathrm{mV}$
Using conservation of linear momentum $P_{i}=P_{f}$
$\therefore 3 \mathrm{~m}=3 \mathrm{~mV}$
$\Rightarrow \mathrm{V}=1 \mathrm{~m} / \mathrm{h}$
.