A bullet of mass 3.9 g was fired at a stationary ball of lead. The head on impact moved the 10 kg ball at a velocity of .3703 m/s. If energy losses due to heat and internal energy are ignored. What was the speed of the bullet?
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.
Let's first determine the momentum of the ball after the collision using the given information. The mass of the ball is 10 kg, and its velocity after the collision is 0.3703 m/s. Therefore, the momentum of the ball after the collision is:
Momentum of the ball = mass of the ball * velocity of the ball
= 10 kg * 0.3703 m/s
= 3.703 kg·m/s
Since there were no other objects involved in the collision besides the bullet and the ball, the total momentum before the collision is equal to the momentum of the bullet. Therefore, we can write:
Momentum of the bullet = Momentum of the ball after the collision
= 3.703 kg·m/s
Now let's calculate the momentum of the bullet using the mass of the bullet (3.9 g = 0.0039 kg) and the speed of the bullet (which we need to find). We can write:
Momentum of the bullet = mass of the bullet * velocity of the bullet
Substituting the given information, we have:
0.0039 kg * velocity of the bullet = 3.703 kg·m/s
Now we can solve for the velocity of the bullet:
velocity of the bullet = 3.703 kg·m/s / 0.0039 kg
≈ 948.97 m/s
Therefore, the speed of the bullet is approximately 948.97 m/s.
If the bullet penetrated the lead and stayed there, then to conserve momentum,
.0039v = (10.0039)(.3703)