91Ó°ÊÓ

Mass of the bullet, m = 3.5 g=0.0035 kg

Mass of the block1,$m_1=1.20\mathrm{kg}$

Mass of the block2$m_2=1.80\mathrm{kg}$

Final speed of the block 1,$v_{1f}=0.630m/s$

Final speed of the block 2,$v_{2f}=1.40m/s$

By applying the law of conservation of momentum forthebullet andthe2^nd block system, find the velocity of the bullet when it is about to enter block 2. This velocity would bethesame as the velocity of the bullet when it leaves block 1. Using this velocity and the velocity of block 1, find the velocity of the bullet when it enters block 1. For this, applythelaw of conservation of momentum to the bullet andthe1^st block system.

Formulae are as follow:

$\begin{array}{l}{P}_i=P_f\\ P=mv\end{array}$

where, m is mass, v is velocity, P is linear momentum.

When the bullet leaves block 1 and enters block 2, the speed of the bullet would bethesame. Assume that the velocity of the bullet when it is about to enter block 2 is. The final velocity of the block and bullet$\left(v_{2f}\right)$isthe same asthe bullet is embedded into the block. It is given as 0.4 m/s.

Now, applythelaw of conservation of momentum to this bullet and 2^nd block system.

So,

$$\begin{gathered} m{v}_{2i}=\left(m+m_2\right)v_{2f} \\ 0.0035×v_i=\left(1.80+0.0035\right)×1.4 \\ {v}_{2i}=721.4m/s \\ ≈721m/s \end{gathered}$$

Hence, the speed of the bullet when it leaves block 1 is 721 m/s.

To calculate the speed of the bullet as it enters block 1, assume the speed of the bullet as $v_{1i}$. The velocity with which the bullet emerges out $\left(v_{2i}\right)$ and the velocity with which block 1 is moving$\left(v_{1f}\right)$ are known. As the masses are known to us, apply the law of conservation of momentum to the bullet and 1^st mass system.

So,

$m{v}_{1i}=m{v}_{2i}+m_1{v}_{1f}$

Substituting the values,

$$\begin{gathered} 0.0035×v_{1f}=\left(0.0035×721.4\right)+\left(1.20×0.630\right) \\ {v}_{1f}=\frac{3.2809}{0.0035} \\ {v}_{1f}=937.4m/s \\ ≈937m/s \end{gathered}$$

Hence,the speed of the bullet when it enters block 1 is 937 m/s.

Therefore, by using law of conservation of momentum, this problem for the velocity of the bullet can be solved.