Question

In Fig. 9-58a, a 3.50 g bullet is fired horizontally at two blocks at rest on a frictionless table. The bullet passes through block 1 (mass 1.20 kg) and embeds itself in block 2 (mass 1.80 kg). The blocks end up with speeds${\mathit{V}}_{\mathbf{1}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{630}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}$and${\mathit{V}}_2\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{40}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}$(Fig. 9-58b). Neglecting the material removed from block 1 by the bullet, Find the speed of the bullet as it (a) leaves and (b) enters block 1.

Short answer

1. Speed of the bullet when it leaves block 1 is 721 m/s
2. Speed of the bullet when it enters block 1 is 937 m/s

Step-by-step solution

Step 1: Given Data

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,$v_{2f}=1.40m/s$

Determining the concept

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.

(a) Determining the speed of the bullet when it leaves block 1

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,

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

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

(b) Determining the speed of the bullet when it enters block 1

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,

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

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.