A ball is shot horizontally from a spring loaded gun into a pendulum catcher (a

Question

A ball is shot horizontally from a spring loaded gun into a pendulum catcher (a metal contrap- tion to catch the ball, hanging from the end of a 25.0 cm long string). The mass of the ball 55.0 g, and the mass of the catcher is 35.0 g. After the collision, the ball+ catcher pendulum rises to an angle of 32.5. a) Is energy conserved after the collision? b) What is the velocity of the pendulum after the collision? c) Is energy conserved during the collision? d) What is the velocity of the ball before it hits the catcher? e) If the gun was only compressed 2.35 cm, what was the spring constant of the spring in the gun?

Explanation / Answer

a)
yes energy is conserved after the collision

b)
Height to which it rises = L - L*cos x
h = 25-25*cos 32.5
= 3.92 cm
= 0.0392 m

use conservation of energy after collision
Kinetic energy just after collision = final potential energy at height h
0.5*(m1+m2)*v^2 = (m1+m2)*g*h
0.5*v^2 = g*h
0.5*v^2 = 9.81*0.0392
v = 0.876 m/s
Answer: 0.876 m/s

c)
No energy is not conserved during collision

d)
Use conservation of momentum,
momentum before collision = momentum after collision
m1*v1 = (m1+m2)*v
55*v1 = (55+35)*0.876
v1= 1.43 m/s
Answer: 1.43 m/s

e)
use conservation of energy
0.5*k*x^2 = 0.5*m1*v1^2
k*x^2 = m1*v1^2
k*(0.0235)^2 = 0.055 * 1.43^2
k = 204.6 N/m
Answer: 204.6 N/m

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