Question

Three guns are aimed at the center of a circle. They are mounted on the circle, $120°$ apart. The fire is in a timed sequence, such that the three bullets collide at the center and mash into a stationary lump. Two of the bullets have identical masses of $4.5g$ each and speeds of $v_1$ and $v_2$. The third bullet has a mass of $2.50g$ and a speed of $575m{s}^{-1}$. Find the unknown speeds.

• A. $200m{s}^{-1}$ each
• B. $145m{s}^{-1}$ and $256m{s}^{-1}$
• C. $536m{s}^{-1}$ and $320m{s}^{-1}$
• D. None of the above

Answer 1

The correct option is C

$536m{s}^{-1}$ and $320m{s}^{-1}$

Step 1: Given Data

As we can see from the figure three bullets are $120°$ apart from each other.

For bullet $2$ and $3$ mass and velocities are identical,

$m_1=m_2=4.50g$

Let $v_1,v_2$ and $v_3$ be the velocities of bullets $1$, $2$, and $3$ respectively.

The mass of bullet 3, $m_2=2.50g$

The velocity of bullet 3, $v_3=575m{s}^{-1}$

Step 2: Formula Used

Let the momentum be,

$p_1+p_2+p_3=0$

According to the law of conservation of momentum,

$m_1{v}_1=m_2{v}_2$

Step 3: Calculate the velocity of bullet $2$

Consider bullet $1$ and $2$, since bullet $3$ has no horizontal component.

$$\begin{gathered} m_1{v}_1\sin 30°=m_2{v}_2\sin 30° \\ \Rightarrow 2.5\times 575\times \sin 30°=450\times v_2\times \sin 30° \\ \Rightarrow v_2=\frac{2.5\times 575}{4.5}=320m{s}^{-1} \end{gathered}$$

Step 4: Calculate the velocity of bullet $3$

Consider vertical components of bullet $1$, $2$, and $3$.

$$\begin{gathered} m_1{v}_1\cos 30°+m_2{v}_2\cos 30°+m_3{v}_3=0 \\ \Rightarrow 2.5\times 575\times \frac{\sqrt{3}}{2}+4.5\times 320\times \frac{\sqrt{3}}{2}+4.5\times v_3=0 \\ \Rightarrow 1244.911+1247.07+4.5\times v_3=0 \\ \Rightarrow 2491.981+4.5\times v_3=0 \\ \Rightarrow 4.5\times v_3=-2491.981 \\ \Rightarrow v_3=\frac{-2491.981}{4.5}=-536m{s}^{-1} \end{gathered}$$

Therefore, the velocity of the third bullet is $536m{s}^{-1}$ in the opposite direction.

Hence, the correct answer is option (C).