the speed of the bullet on impact with the plasticine is v. as a result of the impact, the bullet embeds…

Explanation:

Step1: Apply conservation of mechanical energy for plasticine

Using $mgh=\frac{1}{2}mv^{2}$, where $h = 24\ cm=0.24\ m$, $g = 9.8\ m/s^{2}$. Since we are finding the initial - velocity $v$ of plasticine just after impact and mass $m$ cancels out. $v=\sqrt{2gh}$

Step2: Calculate the value of $v$

Substitute $g = 9.8\ m/s^{2}$ and $h=0.24\ m$ into the formula. $v=\sqrt{2\times9.8\times0.24}\approx2.17\ m/s$

Step3: Apply conservation of momentum for bullet - plasticine system

The initial momentum is $m_{bullet}V$ (where $m_{bullet}=5.2\times 10^{-3}\ kg$ and $V$ is the bullet's initial velocity) and the final momentum is $(m_{bullet}+m_{plasticine})v$. Ignoring the mass of the bullet in part (a) for finding $v$. For part (b), using conservation of momentum $m_{bullet}V=(m_{bullet}+m_{plasticine})v$. We know $m_{bullet}=5.2\times 10^{-3}\ kg$, $m_{plasticine}=0.38\ kg$ and $v\approx2.17\ m/s$. $V=\frac{(m_{bullet}+m_{plasticine})v}{m_{bullet}}$

Step4: Calculate the bullet's initial velocity $V$

$V=\frac{(5.2\times 10^{-3}+ 0.38)\times2.17}{5.2\times 10^{-3}}=\frac{0.3852\times2.17}{5.2\times 10^{-3}}\approx160\ m/s$

Answer:

a. $2.17\ m/s$ b. $160\ m/s$