A 0.02kg bullet is fired at a 3.5kg block which is initially at rest on a table. The bullet embeds in the block. Friction converts the 24.02J of kinetic energy to heat and brings the block and bullet system to stop.
How fast was the bullet and block system sliding immediately after the collision?
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To find the speed of the bullet and block system immediately after the collision, we need to use the principles of conservation of momentum and conservation of kinetic energy.
1. Conservation of momentum:
The total momentum before the collision is equal to the total momentum after the collision. Since the block is initially at rest, the momentum of the bullet and block system before the collision is equal to the momentum of the bullet after the collision.
The momentum (p) of an object is calculated by multiplying its mass (m) by its velocity (v): p = m * v.
Given:
Mass of the bullet (m1) = 0.02 kg
Mass of the block (m2) = 3.5 kg
Let's assume the velocity of the bullet and block system after the collision is V.
Before the collision, the bullet is moving with a certain velocity (v1) and the block is at rest.
Therefore, the momentum before the collision is: p_before = m1 * v1 + m2 * 0 (since the block is at rest, its velocity is 0).
After the collision, the bullet and block move together with the velocity V.
Therefore, the momentum after the collision is: p_after = (m1 + m2) * V.
According to the conservation of momentum, p_before = p_after:
m1 * v1 = (m1 + m2) * V
2. Conservation of kinetic energy:
The total kinetic energy before the collision is equal to the total kinetic energy after the collision. Since the bullet embeds itself in the block, the kinetic energy is converted to heat and no longer present in the system after the collision.
The initial kinetic energy (KE_before) is given by the formula:
KE_before = 0.5 * m1 * v1^2
The final kinetic energy after the collision is zero (as the system has come to rest).
According to the conservation of kinetic energy, KE_before = 0:
0.5 * m1 * v1^2 = 0
To summarize, we have two equations:
1) m1 * v1 = (m1 + m2) * V (from the conservation of momentum)
2) 0.5 * m1 * v1^2 = 0 (from the conservation of kinetic energy)
Solving equation (2) for v1, we get:
0.5 * m1 * v1^2 = 0
v1^2 = 0
v1 = 0
Since the velocity of the bullet before the collision is zero, the bullet and block system is not sliding initially.
Therefore, immediately after the collision, the bullet and block system will also not have any velocity or speed.
In other words, the bullet and block system is at rest after the collision.