Advertisements
Advertisements
Question
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.
Solution 1
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
Solution 2
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`
APPEARS IN
RELATED QUESTIONS
State if the following statement is true or false. Give a reason for your answer.
In an elastic collision of two bodies, the momentum and energy of each body is conserved.
Answer carefully, with reason:
If the potential energy of two billiard balls depends only on the separation distance between their centres, is the collision elastic or inelastic? (Note, we are talking here of potential energy corresponding to the force during collision, not gravitational potential energy.)
The bob A of a pendulum released from 30° to the vertical hits another bob B of the same mass at rest on a table, as shown in the figure. How high does the bob A rise after the collision? Neglect the size of the bobs and assume the collision to be elastic.
A ball is thrown vertically down from height of 80 m from the ground with an initial velocity 'v'. The ball hits the ground, loses `1/6`th of its total mechanical energy, and rebounds back to the same height. If the acceleration due to gravity is 10 ms-2, the value of 'v' is
In Rutherford experiment, for head-on collision of a-particles with a gold nucleus, the impact parameter is ______.
In inelastic collision, ____________.
Two bodies of masses 3 kg and 2 kg collide bead-on. Their relative velocities before and after collision are 20 m/s and 5 m/s respectively. The loss of kinetic energy of the system is ______.
During inelastic collision between two bodies, which of the following quantities always remain conserved?
A cricket ball of mass 150 g moving with a speed of 126 km/h hits at the middle of the bat, held firmly at its position by the batsman. The ball moves straight back to the bowler after hitting the bat. Assuming that collision between ball and bat is completely elastic and the two remain in contact for 0.001s, the force that the batsman had to apply to hold the bat firmly at its place would be ______.
A sphere of mass 'm' moving with velocity 'v' collides head-on another sphere of same mass which is at rest. The ratio of final velocity of second sphere to the initial velocity of the first sphere is ______. ( e is coefficient of restitution and collision is inelastic)
