Question

A bullet of mass 10g travelling horizontally with a velocity of 150m/s strikes a stationary wooden block.also calculate the magnitude of the force extered by the wooden block on the bullet.

Answer 1

AnswEr:

$\bf{Given}\begin{cases}\sf{Mass,m=10g.=0.01kg}\\\sf{Initial\:Velocity,u=150m/s}\\ \sf{Final\:velocity,v=0}\\\sf{Time,t=0.03s} \end{cases}$

• $\sf\underline\pink{\:\:\:\:\:Retardation:-\:\:\:\:}$

$\longrightarrow \sf \frac{v - u}{t} \\ \\ \longrightarrow \sf \: \frac{0 - 150}{0.03} \\ \\ \longrightarrow \sf \: - 5000ms - {}^{2}$

Force = m × a

$\longrightarrow \sf \: 0.01 \times - 5000 \\ \\ \longrightarrow \sf \: 50 \: n$

• $\sf\underline\blue{\:\:\:Distance\:of\:penetrations:-\:\:\:\:}$

$\longrightarrow \sf \: \frac{ {v}^{2} - {u}^{2} }{2a} \\ \\ \longrightarrow \sf \frac{ {0}^{2} - {(150)}^{2} }{2 \times ( - 5000)} \\ \\ \longrightarrow \sf \: 2.25 \: m$

Answer 2

It is given that the mass of of the bullet is 0.01 kg, initial and final velocities are 150 m/s and 0 m/s respectively. (And time taken is 0.03 seconds.)

We have to find the force (F) applied by the block on the bullet.

It is known that,

$\frak{\textcolor{red} {F = ma}}$ and

$\frak{ \textcolor{red}{a = \frac{v-u}{t}}}$

where,

• F = Force,
• m = Mass,
• a = Acceleration
• v, u = Final and initial velocities respectively and
• t = Time.

Or,

$\frak{ \textcolor{red}{F = m (\frac{v-u}{t})} }$

or, $\frak{\textcolor{red}{F = m (\frac{-u}{t})} }$

(In this case only.)

↗ F = (0.01 kg)[(- 150 m/s)/(0.03 s)]

↗ F = (0.01 kg)[(- 150 m/s)(100)/3 s]

↗ F = (0.01 kg)(- 50 m/s²)(100)

↗ F = (1 kg)(- 50 m/s²)

↗ F = - 50 kg m/s²

$\frak{\textcolor{red}{\therefore F = - 50 N} }$

(On the direction of motion)

Hence, the force provided by the block was of 50 newtons.