Question

A bullet of mass 30g is fired with a speed of 400m/s into a sandbag. The sandbag has a mass of 10kg and its suspeneded by two ropes so that it can swing .what is the maximum vertical height ,that the sandbag rises as it recoiled with the bullet logged in ?

Answer 1

$\huge\mathbb{\underline{Answer:-}}$

Mass of bullet(m)= 30g = 0.03kg.

Initial velocity (u)= 400m/s.

Final velocity (v)= 0m/s. [Since bullet comes to rest after penetrating 10 cm in wall]

Distance penetrated by bullet (S) or displacement of bullet in presence of resistive force = 10 cm = 0.1m.

Acceleration (a)=?

From third equation of motion:

v^2=u^2+2aS

0^2= 400^2+2a*0.1

0.2a=-160000

a=-160000/0.2= -8,00,000m/s^2.

The negative sign shows that acceleration takes place in direction opposite to the initial direction of motion of bullet or retardation takes place.

Avg. force (f)= m*a

f= 0.03*-800000

f= -24000N

Here too negative sign shows that force applied by mud wall on bullet is in direction opposite to initial direction of motion of bullet.

Answer 2

Answer:

conserving the momentum of system.

0.025∗400+4.975∗0.0

=(4.975+0.025)∗v

v=10/5=2m/s