Question

A bullet with speed 511 m/s and mass m = 5.00 g is shot into a stationary block of wood hanging a distance
h1 = 1.00 m above the ground, as shown in Figure 1. The block is hanging by a string of length r = 0.30 m. The
bullet is lodged in the block, and the block-bullet system swings so that the string makes an angle of θ = 60.0°
when the string breaks, as shown in Figure 2. The block-bullet system then falls to the ground a horizontal
distance d2 = 2.00 m from its position when the string breaks. Ignoring the effects of air resistance, determine
the mass, M of the block of wood.

Answer 1

Explanations & Calculations

• Refer to the figure & consider the mass of the block to be M kg.
• Applying the conservation of linear momentum to find V (velocity of the system after the collision)

$\qquad\qquad \begin{aligned} \small 0.005kg\times \overrightarrow{511} ms^{-1} +0 &=\small (0.005 + M) V\\ \small V &= \small \frac{2.555}{(0.005 +M)} \end{aligned}$

• From down here to up where the line brakes, apply conservation of mechanical energy to find the velocity effective for the projectile.

(Here the height where the line breaks = 1+h = $1+(\small 0.3 -0.3\cos60) = \small \bold{1.15m}$ )

$\qquad\qquad \begin{aligned} \small E_k +E_p &= \small E_{k1} +E_{p1}\\ \scriptsize \frac{1}{2}(0.005+M)\times\frac{2.555^2}{(0.005+M)^2} +0 &= \scriptsize\frac{1}{2}(0.005+M)V_1^2 + (0.005+M)gh\\ \small \frac{2.555^2}{2(0.005+M)} &= \small \frac{1}{2}(0.005+M)(V^2+2gh)\\ \small (0.005+M)^2 &= \small \frac{2.555^2}{(V^2+2gh)} \cdots(1) \end{aligned}$

• Now to find V, apply s = ut + 0.5at² both horizontally & vertically down.

$\qquad\qquad \begin{aligned} \small s &= ut\\ 2 &= \small V\cos60\times t\\ \small t&= \small\frac{4}{V}\cdots(2) \end{aligned}$ $\qquad\qquad \begin{aligned} \small \downarrow s &= ut+0.5gt^2\\ \small 1.15 &= \small -V\sin60\times t+0.5\times9.8\times t^2\\ \small 1.15 &= \small \frac{-\sqrt{3}Vt}{2}+0.5\times9.8\times t^2\\ \end{aligned}$

• Now solving for V yields,

$\qquad\qquad \small V=\begin{cases} 4.122 ms^{-1}\\ -4.122 ms^{-1} &\text{negligible; negative} \end{cases}$

• Now apply this value in (1) to calculate M.

$\qquad\qquad \begin{aligned} \small (0.005+M)^2 &= \small \frac{2.555^2}{(V^2+2gh)}\\ \small (0.005+M)^2 &= \small \frac{2.555^2}{\bold{4.122}^2+2\times9.8 \times0.15}\\ \end{aligned}$

• Solving for M yields,

$\qquad\qquad \small M=\begin{cases} 0.567kg\\ -0.577kg &\text{negligible; negative} \end{cases}$

• Therefore, the mass of the block is M = 0.567 kg