Answer:
V=0.28m/s, the combined mass will be 0.28m/s fast
h=0.00399m
Explanation:
Mass m is fired to the right at a speed of 7.0 m/s where it collides and becomes imbedded in mass M which is hanging from a rope. Mass m is 0.50kg and mass M is 12.0kg. a- How fast will the combined mass (m+M) be moving just after the collision? b- What is the maximum height the system (m+M) will reach?
step by step explanation:
from the principles of linear momentum which states that the total momentum before collision is equal to the total momentum after collision
momentum before collision=momentum after collision
notice that in this this case the collision is inelastic, owing to the fact that the two bodies stuck together after collision
mu+Mu2=(m+M)V
V=common velocity after collision
m=mass of the body fired
M=Mass hanging on the rope
u=velocity of the body
u2=velocity of the mass hanging on arope 0m/s
0.5*7+0=(0.5+12)V
3.5=12.5V
V=3.5/12.5
V=0.28m/s, the combined mass will be 0.28m/s fast
b. the maximum height reached after the impact
starting with the swing of the pendulum just after the collision until it reaches its maximum height. Conservation of mechanical energy applies here, so:
kinetic energy =potential energy
½ (m+M)V = (m+M)gh.
So, the speed of the system immediately after the collision is:
V = (2gh)½
0.28=(2*9.81*h)^0.5
0.0784=2*9.81h
h=0.00399m
