Question

A bullet with initial speed 500 m/s passes straight through a pendulum bob of mass 2 kg suspended by a 2 m long string. The bullet emerges from the bob with a speed of 100 m/s. Find mass of bullet if, after impact, the bob swings through a complete vertical circle with speed 6 m/s in the top point of the circle.

Answer 1

Answer 2

Step 1: Verify the minimum speed for the vertical circle

The minimum speed at the top of the circle is $\sqrt{gL}=\sqrt{9.8\times 2}\approx 4.43\text{m/s}$$4.43\text{m/s}$. The given $v_{\text{top}}=6\text{m/s}$$v_{\text{top}}=6\text{m/s}$ satisfies this condition.

Step 2: Relate $v_{\text{bottom}}$ to $v_{\text{top}}$ using energy conservation

$$\frac{1}{2}M{v}_{\text{bottom}}^2=\frac{1}{2}M{v}_{\text{top}}^2+Mg(2L)$$

Solving for $v_{\text{bottom}}$:

$$v_{\text{bottom}}=\sqrt{v_{\text{top}}^2+4gL}=\sqrt{6^2+4\times 9.8\times 2}=\sqrt{114.4}\approx 10.7\text{m/s}\mathrm{.}$$

Step 3: Apply conservation of momentum

$$m{v}_{1i}=m{v}_{1f}+M{v}_{\text{bottom}}$$

Substitute values:

$$500m=100m+2\times 10.7⟹400m=21.4⟹m=\frac{21.4}{400}=0.0535\text{kg}=53.5\text{g}\mathrm{.}$$

Answer is :

The mass of the bullet is $\boxed{{\displaystyle 53.5\text{g}}}$