Question

A bullet of mass 50 g moving with an initial velocity of 100 m/s strikes a wooden block and comes to rest after penetrating a distance of 2 cm in it. Calculate
(i) initial momentum of a bullet
(ii) final momentum of a bullet
(iii) retardation caused by the wooden block and
(iv) resistive force exerted by the wooden block

Answer 1

Here we have been given with mass of the object, its initial and final velocities. The momentum of the body (bullet) can be calculated using the formula of momentum: $p=mv$. Then by using the formula: ${{v}^{2}}={{u}^{2}}+2as$, calculate the acceleration of the bullet for covering a distance of 2 cm. Then, by using the formula: $F=ma$ find the resistive force exerted by the wooden block.

Formula used:
$p=mv$, where p is momentum, m is mass and v is the velocity
${{v}^{2}}={{u}^{2}}+2as$, where v is final velocity, u is initial velocity, a is acceleration and s is the distance travelled.
$F=ma$, where F is force, m is the mass and a is the acceleration.

Complete step by step answer:
We have:
$\begin{aligned} & m=50g \\\ & u=100m/s \\\ & v=0m/s \\\ & s=2cm \\\ \end{aligned}$
As we know that, momentum is given as: $p=mv$
(i) The initial momentum of bullet is:

& {{p}_{i}}=0.050\times 100 \\\ & =5kgm/s \end{aligned}$$ (ii) The final momentum of bullet is: $\begin{aligned} & {{p}_{f}}=0.05\times 0 \\\ & =0kgm/s \end{aligned}$ As we know that: ${{v}^{2}}={{u}^{2}}+2as$ So, we have: $\begin{aligned} & \Rightarrow 0={{\left( 100 \right)}^{2}}+2\times a\times \dfrac{2}{100} \\\ & \Rightarrow a=-250000m{{s}^{-2}} \\\ \end{aligned}$ (ii) The retardation caused by the wooden block is $250000m{{s}^{-2}}$ Now, we know that: $F=ma$ (iv) The resistive force on the bullet is: $\begin{aligned} & F=0.05\times -250000 \\\ & =-12500N \end{aligned}$ . **Note:** Momentum of any particle is the product of mass and its velocity. Therefore, any particle having a large velocity will also have large momentum. Momentum is a vector quantity. This means that it has both magnitude and direction.