A ball of mass 0.075 kg is fired horizontally into a ballistic pendulum. The pendulum mass is 0.350 kg. The ball is caught in the pendulum, and the centerof mass of the system rises a vertical distance of 0.145 m in the earth's gravitational field. What was the original speed of the ball? Assume g=9.80 m/s^2.
To find the original speed of the ball, we can use the principle of conservation of energy and momentum.
First, let's calculate the change in potential energy of the system. The ball rises a vertical distance of 0.145 m, so the change in potential energy is:
ΔPE = mgh
= (0.075 kg + 0.350 kg) * 9.80 m/s^2 * 0.145 m
= 0.0771 J
Since the initial kinetic energy of the system is equal to the change in potential energy, we can write:
ΔKE = 0.0771 J
Now, let's calculate the initial kinetic energy of the ball. The kinetic energy of an object is given by:
KE = 0.5 * mv^2
where m is the mass of the ball and v is its velocity. Rearranging this equation, we have:
v^2 = (2 * KE) / m
Plugging in the values, we get:
v^2 = (2 * 0.0771 J) / 0.075 kg
= 2.052 J/kg
Now, solving for v:
v = √(2.052 J/kg)
= 1.43 m/s
Therefore, the original speed of the ball was approximately 1.43 m/s.