1. A 0.80 kg arrow flying in a straight line at 70.0 m/s hits and sticks in a bl
ID: 2250918 • Letter: 1
Question
1. A 0.80 kg arrow flying in a straight line at 70.0 m/s hits and sticks in a block of wood at rest. If the mass of wood block is 5.4 kg, what is the velocity of the two after the collision?
2. What is
a) the total kinetic energy of the arrow and block of wood in Question #1 before the collision?
b) the total kinetic energy of the arrow and block of wood in Question #1 after the collision?
c) Is the initial KE equal to the final KE? If not, explain your answer in respect to the Law of Conservation of Energy.
d) the name of this type of collision?
3. A red pool ball of mass 0.12 kg is moving to the right at a speed of 5.0 m/s. It collides head-on with a black pool ball whose mass is the same and is moving to the left at 2.5 m/s. After the collision, the red ball is moving to the left at a speed of 2.5 m/s. What is the speed of the black ball after the collision?
4. What is
a) the total kinetic energy of the red ball and black ball in Question #3 before the collision?
b) the total kinetic energy of the red ball and black ball in Question #3 after the collision?
c) Is the initial KE equal to the final KE? If not, explain your answer in respect to the Law of Conservation of Energy.
d) the name of this type of collision?
Explanation / Answer
1) conservation of momentum: 0.8(70) + 5.4 (0) = (5.4+0.8)v
Velocity of the two after collision is 9.032 m/s
2)Total KE before collision = 0.5*0.8*70^2 = 1960 J
Total KE after collision = 0.5*(5.4+0.8)*9.032^2 = 253 J
No. KE before is not equal to that after. Energy is lost in the process of collision as heat.
Inelastic collision
3)conservation of momentum: 0.12(5) + 0.12(-2.5) = 0.12(-2.5) + 0.12v
speed of the black ball after the collision is 5 m/s towards right
4)KE before collision = 0.5*0.12*(5^2+2.5^2) =1.875 J
KE after collision = 0.5*0.12*(2.5^2+5^2) =1.875 J
YES. KE before collision =KE after collision
ELASTIC collision
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