Question

A bullet of mass 0.05kg travelling horizontally at 30ms^-1 embeds itself in wooden block of mass 0.95kg.if the block is suspended by a light vertical string 1m long calculate the maximum inclination of the string at the horizontal

Answer 1

Explanations & Calculations

• The finding of the speed of the block + bullet combined system with the help of conservation of linear momentum.

$\qquad\qquad \begin{aligned} \small 0.05kg\times30ms^{-1}+0&=\small (0.05kg+0 .95kg)V\\ \small V&=\small 1.5\,ms^{-1} \end{aligned}$

• This kinetic energy is expended as the block-bullet system moves up to the topmost position possible.
• If the inclination set up with the vertical is $\small \theta,$ then the height to the block-bullet system from the ground is

$\qquad\qquad \begin{aligned} \small h&=\small L-L\cos\theta\\ &=\small 1(1-\cos\theta) \end{aligned}$

• Finally, by conservation of mechanical energy,

$\qquad\qquad \begin{aligned} \small \frac{1}{2}mV^2+0&=\small 0+(m+M)gh\\ \small \frac{1}{2}. 0.05kg.(1.5ms^{-1})^2&=\small (0.05+0.95).9.8.(1-\cos\theta)\\ \small 1-\cos\theta&=\small 0.00576\\ \small \theta&=\small 6.12^0 \end{aligned}$

• Therefore, the inclination measured w.r.t the horizontal is

$\qquad\qquad\qquad\small 90-6.12 =83.88^0$