Question

2. A bullet of mass 0.012 kg and horizontal speed 70 ms
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. 1
a. 72.2 cm, 28.5 J
b. 21.2 cm, 28.5 J
c. 29.2 cm, 29.5 J
d. 23.2 cm, 38.5 J
3. A rain drop of mass 2.00 g falls from a height 1.00 km. It
hits the ground with a speed of 50.0 m/s. What is the
work done by the resistive forces if any? (Take g = 10 m
s-2) 1
a. - 27.5 J
b. - 22.5 J
C. - 19.5 J
d. - 17.5 J​

Answer 1

Answer:

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

Explanation:

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