A bullet of mass 0.08kg is fired horizontally into a 10kg block which is free to move. If both bullet and block move with a velocity of 0.6m/s after the impact, find the velocity with which the bullet hits the body
The law of conservation of momentum applies
mV+0=(M+m)V'
solve for V
To solve this problem, we can apply the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.
The momentum (p) is defined as the product of mass (m) and velocity (v). Mathematically, p = mv.
Given information:
Mass of the bullet (mb) = 0.08 kg
Mass of the block (mblk) = 10 kg
Velocity after the impact = 0.6 m/s (common velocity for both bullet and block)
Before the impact, the bullet is the only object in motion, so its momentum is mb * vb, where vb is the velocity with which the bullet hits the block.
After the impact, both the bullet and block are moving together with a common velocity (0.6 m/s), so their combined momentum is (mb + mblk) * 0.6.
Applying the principle of conservation of momentum:
mb * vb = (mb + mblk) * 0.6
Now, let's solve for vb:
0.08 kg * vb = (0.08 kg + 10 kg) * 0.6
0.08 kg * vb = 10.48 kg * 0.6
0.08 kg * vb = 6.29 kg m/s
Divide both sides by 0.08 kg to isolate vb:
vb = 6.29 kg m/s / 0.08 kg
vb ≈ 78.63 m/s
Therefore, the velocity with which the bullet hits the block is approximately 78.63 m/s.