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प्रश्न
A bullet of mass 0.012 kg and horizontal speed 70 m s–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.
उत्तर १
Initial speed of the bullet, ub = 70 m/s
Mass of the wooden block, M = 0.4 kg
Initial speed of the wooden block, uB = 0
Final speed of the system of the bullet and the block = ν
Applying the law of conservation of momentum:
mub + MuB = (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 = 2.04 m/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'v2
∴ h = (1/2)(v2 / g)
= (1/2) × (2.04)2 / 9.8
= 0.2123 m
The wooden block will rise to a height of 0.2123 m.
Heat produced = Kinetic energy of the bullet – Kinetic energy of the system
= (1/2) mu2 - (1/2) m'v2 = (1/2) × 0.012 × (70)2 - (1/2) × 0.412 × (2.04)2
= 29.4 - 0.857 = 28.54 J
उत्तर २
Here, m1 = 0.012 kg, u1 = 70 m/s
m2 = 0.4 kg, u2 = 0
As the bullet comes to rest with respect to the block, the two behave as one body. Let v be the velocity acquired by the combination.
Applying principle of conservation of linear momentum, (m1 + m2) v = m1H1 + m2u2 = m1u1
`v = (m_1u_1)/(m_1+m_2) = (0.012xx70)/(0.012+0.4) = (0.84)/0.412 = 2.04 ms^(-1)`
Let the block rise to a height h
P.E. of the combination = K.E. of the combination
`(m_1+m_2)gh = 1/2(m_1+m_2)v^2`
`h = v^2/(2g) = (2.04xx2.04)/(2xx9.8) = 0.212 m`
For calculating heat produced, we calculate energy lost (W), where
W = intial K.E. of bullet - final K.E of combination
`= 1/2m_1u_1^2 - 1/2(m_1+m_2)v^2`
`1/2xx0.012(70)^2 - 1/2(0.412)(2.04)`
W = 29.4 - 0.86 = 28.54 joule
:. Heat produced `H = W/J = (28.54)/4.2 = 6.8 cal`
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