Question

The collision between a geological hammer and a rock lying loose on the ground can be considered to be approximately elastic.

Calculate the final speed of a 0.21-kg rock when it is struck by a 0.55-kg hammer moving with an initial speed of 5.2 m/s. The rock is initially at rest.

Answer 1

The final velocity of the rock when struck by the hammer can be calculated using the principles of conservation of momentum and kinetic energy.

Step-by-step explanation:

When a hammer collides with a rock, the two objects experience a collision. In this case, the collision is considered to be approximately elastic, which means that both momentum and kinetic energy are conserved.

1. First, calculate the initial momentum of the hammer. This can be done by multiplying its mass (0.55 kg) by its initial velocity (5.2 m/s). The initial momentum of the hammer is 2.86 kg•m/s.
2. Since the rock is initially at rest, its initial velocity is 0 m/s and its initial momentum is 0 kg•m/s.
3. Next, use the conservation of momentum principle to determine the final momentum of the system. The final momentum of the hammer and the rock should be equal to the initial momentum of the hammer. Therefore, the final momentum of the system is also 2.86 kg•m/s.
4. Finally, calculate the final velocity of the rock by dividing the final momentum of the system by the mass of the rock (0.21 kg). The final velocity of the rock is approximately 13.6 m/s.

Answer 2

given,

mass of the rock ( m ) = 0.21 Kg

mass of hammer ( M ) = 0.55 Kg

initial speed of the hammer ( u ) = 5.2 m/s

final speed of the rock = ?

using energy of conservation

`Mu = M v + m v'`.....(1)

In elastic collision Kinetic energy is equal to final energy

`(1)/(2)Mu^2 = (1)/(2)Mv^2 + (1)/(2)Mv'^2`....(2)

on solving both the equation for velocity of hammer

`v' = (M-m)/(M+m)u`

`v' = (0.55-0.21)/(0.55+0.21)* 5.2`

v' = 2.33 m/s

velocity of the rock

`v = (2M)/(M+m)v'`

`v = (2* 0.55)/(0.55 + 0.21)* 2.33`

v = 3.37 m/s