In a ballistic pendulum, an object of mass m is fired with an initial speed v i
ID: 1547007 • Letter: I
Question
In a ballistic pendulum, an object of mass m is fired with an initial speed vi at the bob of a pendulum. The bob has a mass M, and is suspended by a rod of negligible mass. After the collision, the object and the bob stick together and swing through an arc, eventually gaining a height h.
A) Suppose a bullet of mass
m = 6.95 kg is fired into a ballistic pendulum whose bob has a mass of M = 0.650 kg. If the bob rises to a height of 0.148 m, what was the initial speed of the bullet?
B) What was the speed of the bullet-bob combination immediately after the collision took place?
C)Refer back to example 9-12. A bullet with a mass of
m=8.10g and an initial speed of vi=320m/s is fired into a ballistic pendulum. What mass must the bob have if the bullet-bob combination is to rise to a maximum height of 0.125 m after the collision?
Explanation / Answer
Using law of conservation of energy
Energy of the system before collision = energy of the system after the collision
0.5*(m+M)*V^2 = (m+M)*g*h
0.5*((6.95*10^-3)+(0.65))*V^2 = (0.00695+0.65)*9.8*0.148
speed of the bob - bullet immediately after collision is V = 1.703 m/sec
then
using law of conservation of momentum
momentum before collision = momentum after collision
(m*Vo) = (m+M)*V
(6.95*10^-3*Vo) = (0.00695+0.65)*1.703
Vo = 160.97 m/sec is the speed of the bullet
B) speed of the bob - bullet immediately after collision is V = 1.703 m/sec
C) using law of conservation of momentum
momentum before collision = momentum after collision
(m*Vo) = (m+M)*V
(8.1*10^-3*320) = (0.0081+M)*V
2.592 = (0.0081+M)*V
speed of the bob and bullet immidiately after the collision is V = 2.592/(0.0081+M)
Using law of conservation of energy
Energy of the system before collision = energy of the system after the collision
0.5*(m+M)*V^2 = (m+M)*g*h
0.5*((8.1*10^-3)+M)*((2.592/(0.0081+M))^2 = (0.0081+M)*9.8*0.125
0.5*2.592 = (0.0081+M)^2*9.8*0.125
Mass of the bob is M = 1.02 Kg