Question

a body of mass 5 kg resting on a horizontal surface. A bullet of mass 200 gtam strikes it with a speed 300m/sec and embedded with it combined mass moves a distance 400m .Find the value of coffiecent og friction​

Answer 1

Explanation:

Mass of the bullet, m = 0.012 kg

Initial speed of the bullet, u

b

=70m/s

Mass of the wooden block, M=0.4 kg

Initial speed of the wooden block, u

B

=0

Final speed of the system of the bullet and the block = v m/s

Applying the law of conservation of momentum:

mu

b

+Mu

B

=(m+M)v

0.012×70+0.4×0=(0.012+0.4)v

v=0.84/0.412

=2.04 m/s

For the system of the bullet and the wooden block:

Mass of the system, m

′

=0.412 kg

Velocity of the system =2.04m/s

Height up to which the system rises = h

Applying the law of conservation of energy to this system:

Potential energy at the highest point = Kinetic energy at the lowest point

m

′

gh=(1/2)m

′

v

2

h=

2g

v

2

=

2×9.8

(2.04)

2

=0.2123m

The wooden block will rise to a height of 0.2123m.

The heat produced = Kinetic energy of the bullet - Kinetic energy of the system

=(1/2)mu

2

−(1/2)m

′

v

2

=(1/2)×0.012×(70)

2

−(1/2)×0.412×(2.04)

2

=29.4−0.857=28.54J