a bullet of mass 5.0g was fired at a stationary ball of lead. the head on impact moved the 10 kg ball at a velocity of .50m/s. of energy losses due heat the internal energy are ignored, what was the speed of the bullet?
You will use the conservation of momentum for this question.
Mass of bullet = m = 0.005kg
Mass of ball of lead = M = 10kg
Initial speed of bullet = u
Final combined speed = v = 0.5m/s
As per conservation of momentum,
mu = (M+m)v
=> (0.005)u = (0.005+10)(0.5)
=> u = (10.005)(100)
=> u = 1000.5 m/s
Hence, the original speed was 1000.5 m/s
To find the speed of the bullet, we can use 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.
Before the collision:
The bullet is moving with an unknown velocity, let's call it "vb."
The ball is stationary, so its velocity, "vb" = 0.
After the collision:
The bullet is embedded into the ball. Let's call the final velocity of the combined bullet and ball "vf."
The bullet and ball move together, so their final velocity, "vf" = 0.50 m/s.
Now, we can use the principle of conservation of momentum to solve for the initial velocity of the bullet.
Momentum before = momentum after
(mass of bullet) x (velocity of bullet before) = (mass of bullet + mass of ball) x (velocity after)
Converting the mass of the bullet from grams to kilograms:
mass of bullet = 5.0 grams = 0.005 kg
Plugging in the values:
(0.005 kg) x (vb) = (0.005 kg + 10 kg) x (0.50 m/s)
0.005 vb = (0.005 + 10) x 0.50
0.005 vb = (10.005) x 0.50
Now, we can solve for vb:
vb = (10.005 x 0.50) / 0.005
vb = 5.0025 / 0.005
vb = 1000.5 m/s
Therefore, the speed of the bullet before the collision was approximately 1000.5 m/s.