Ballistic Pendulum Question

Question

A 0.02 kg bullet is fired into a 2.4 kg sandbag suspended by a wire of negligible mass. The entire system swings up to a height of 0.84 m. If the initial speed of the bullet is 490 m/s, what is the loss in mechanical energy due to the collision?

Answer 1

Subtract the potential energy of the system at maximum height from the initial kinetic energy of the bullet.  That will be the amount of mechanical energy that is lost.  

 

m = mass of bullet = 0.02 kg

v = initial speed of bullet = 490 m/s

h = distance the sandbag + bullet system rises = 0.84 m 

M = mass of sandbag = 2.4 kg

ΔE = energy lost = ?

 

ΔE = mv²/2 - (m + M)gh = [(0.02)(490)²/2 - (0.02 + 2.4)(9.8)(0.84)] J = (2401 - 19.92) J ≅  2381 J

 

Answer 2

The information about the swing up to 0.84 m is not really needed in this version of the ballistic pendulum problem.

 

In the collision, momentum is conserved, but not energy.

 

To use the principle of conservation of momentum, write  pi = pf

pi =   0.02 x 490 = 9.8

pf =  (2.4 + .02) vf  ;   forcing pi = pf  results in vf =  4.05  

 

The energy before the collision is (1/2) .02  490^2 =  2401 J   

The energy after is (1/2) (2.4 + .02)  4.05^2 =   19.85

Thus the loss of energy is 2401 - 19.85 =  2384  J