Question

A bullet of mass 0,02 kg is fired horizontally into a suspended stationary wooden block of mass 1,95 kg with a velocity of 20 m-s1, as shown in the diagram. Ignore the effects of air resistance. 2.1 Define, in words, the term momentum as applied in physics. 2.2 Calculate the momentum of the bullet just before impact. The bullet is embedded in the wooden block after collision. 2.3 State the principle of conservation of linear momentum in words. 2.4 Calculate the common velocity of both the bullet and the block after collision 2.5 Is this collision ELASTIC or INELASTIC? Use appropriate calculations to explain the answer​

Answer 1

28.54J

Step-by-step explanation:

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Class 11

>>Physics

>>Systems of Particles and Rotational Motion

>>Linear Momentum of a System of Particles

>>A bullet of mass 0.012 kg and horizontal

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A bullet of mass 0.012 kg and horizontal speed 70 ms

−1

strikes a block of wood of mass 0.4 kg and instantly comes to rest with respect to the block. The block is suspended from the ceiling by means of thin wires. Calculate the height to which the block rises. Also, estimate the amount of heat produced in the block.

Hard

Solution

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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