Question

A 3.00 kg box sliding west at 2.00 m/s makes an inelastic collision with a second box sliding 1.50 m/s east. afterwards, they both come to a stop. what was the mass of the second box?

Answer 1

Here,

• Mass of block moving westward, $\sf{m_1=3\ kg}$

Let mass of the other block be $\sf{m_2.}$

We consider eastward direction as positive.

• Initial velocity of block moving westward, $\sf{u_1=-2\ m\,s^{-1}}$

• Initial velocity of block moving eastward, $\sf{u_2=1.5\ m\,s^{-1}}$

Here the blocks come to rest after collision, so their final velocities will be zero.

By law of conservation of linear momentum,

$\longrightarrow\sf{m_1u_1+m_2u_2=(m_1+m_2)0}$

$\longrightarrow\sf{m_2u_2=-m_1u_1}$

$\longrightarrow\sf{m_2=-\dfrac{m_1u_1}{u_2}}$

$\longrightarrow\sf{m_2=-\dfrac{3\times-2}{1.5}}$

$\longrightarrow\sf{\underline{\underline{m_2=4\ kg}}}$

Hence mass of second block is $\bf{4\ kg.}$