13-The mass of the blue puck shown below is 25.0% greater than the mass of the g
ID: 2244409 • Letter: 1
Question
13-The mass of the blue puck shown below is 25.0% greater than the mass of the green puck. Before colliding, the pucks approach each other with momenta of equal magnitudes and opposite directions, and the green puck has an initial speed of 10.0 m/s. Find the speeds the pucks have after the collision if half the kinetic energy of the system becomes internal energy during the collision.
Explanation / Answer
Let m1 = mass of green puck
m2 = mass of blue puck
u1 = initial speed of green puck
u2 = initial speed of blue puck
m2 = m1 + 25% of m1
m2 = 1.25 m1--------------(1)
u1 = 10.0 m/s------------(2)
Before collision, the pucks momenta are equal and opposite.
Therefore total momentum = 0
m1 u1 = m2 u2
m1 * 10.0 = 1.25m1 * u2------------[using (1) and (2)]
10.0 = 1.25 * u2
u2 = 10.0/1.25
u2 = 8.0 m/s---------(3)
Total KE before collision = (1/2)m1 u1^2 + (1/2)m2 u2^2
= (1/2)m1 * 10^2 + (1/2) * 1.2 5m1 * 8^2
= 50 m1 + 40 m1
= 90 m1
Half KE is converted into internal energy.
Therefore total KE after collision = (90/2)m1 = 45 m1-----(4)
Let speed of green puck after collision = v1
speed of blue puck after collision = v2
By conservation of momentum,
m1 v1 = m2 v2
m1 v1 = 1.2 m1 v2
v1 = 1.2 v2-----------------(5)
Total KE after collision
= (1/2)m1 v1^2 + (1/2)m2 v2^2
Using (1), (4), (5) in the above,
45m1 = (1/2)m1 (1.25 v2)^2 + (1/2) * 1.25m1 * v2^2
45= (1/2)(1.25 v2)^2 + (1/2) * 1.2 5* v2^2
66 = 0.78 v2^2 + 0.62 v2^2
66 = 1.4 v2^2
v2^2 = 45/1.4 = 32.14
v2 = sqrt(50) = 5.66m/s
From (5),
v1 = 1.2 5* v2 = 1.25 * 5.66 = 7.086 m/s
Ans: Speed of green puck = 7.086m/s
Speed of blue puck = 5.66 m/s
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