The following information applies to the next four questions (36 through 39) As
ID: 1508805 • Letter: T
Question
The following information applies to the next four questions (36 through 39) As demonstrated in class, using a ballistic pendulum to capture a ping-pong ball that is fired out of an air can clever way to de ong ball that is fired out of an air cannon is a cever way to determine the muzzle speed of the ping-pong ball. The figure below illustrates the set up. Mass ofcan catching ball, M= 0.400 kg Mass of ping-pong ball, m 0.00250 kg Height of pendulum swing, h - 0.190 m Ping-Pong Ball Cannon IN Ballistic Pendulum The ball is fired out of the cannon, where it is trapped by a can hanging from a string that captures the ball. The can is initially at rest. After the ball impacts, and is trapped by the can, the can and the ball swing together to a height h above the initial starting height of the can. The mass of the ping-pong ball is m = 0.00250 kg. The mass of the can is M = 0.400 kg. The height to which the can of the ballistic pendulum swings is h = 0.190 meters.Explanation / Answer
Because the collision within the ball-can system is so brief, we can make two important assumptions: (1) During the collision, the gravitational force on the can and the force on the can from the cords are still balanced. Thus, during the collision, the net external impulse on the ball-can system is zero. Therefore, the system is isolated and its total linear momentum is conserved:
that is: total momentum before collision = total momentum after collision
which implies V = [m/(m + M)]*v (1)
as the can and the ball now swing together, the total mechanical energy of the ball-earth-can system is conserved, that is:
mechanical energy at bottom = mechanical energy at top
therefore 1/2(m + M) V2 = (m + M)gh (2)
susbtituing for V from equation (1) into (2), we get
v = [(m + M)/m] * sqrt(2gh)
now plugging in the values, we get
v = [(0.0025 + 0.4)/0.0025] * sqrt (2 * 9.8 * 0.19) = 310.7m/s
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