The drawing shows a collision between two pucks on an air-hockey table. Puck a m
ID: 1523336 • Letter: T
Question
The drawing shows a collision between two pucks on an air-hockey table. Puck a mass 0.065 kg and is moving along the x axis with a velocity of +5.5 m/s. It makes a collision with puck B, which has a mass of 0.040 kg and is initially at rest. The collision is not head-on. After the collision, the two pucks fly apart with the angles shown in the drawing. Write an equation that describes the momentum of the system before and after the collision in the x-direction. Write an equation that describes the momentum of the system before and after the collision in the y-direction. Find the final speed of puck A. Find the final speed of puck B.Explanation / Answer
m1= 0.065kg
m2= 0.040kg
The collision is two dimensional collision. The momentum is conserved along the horizontal and also perpendicular to the direction of collision.
along the direction of collision
m1 u1 + 0 = m1 V1 cos 65 + m2 V2 cos 37 ......(1)
(0.065)(5.5) = (0.065) V1 cos 65 + (0.040) V2 cos 37......(2)
Momentum perpendicular to the direction of impact is
0 + 0 = m1 V1 sin 65 - m2 V2 sin 37 .......(3)
m1 V1 sin 65 = m2 V2 sin 37
V1 / V2 = m2 sin 37 / m1 sin 65
= (0.040) sin 37 / (0.065) sin 65
= 0.40863
V1 = 0.40863 V2 .......(4)
Substituting equation (4) in (2)
(0.065)(5.5) = (0.065) V1 cos 65 + (0.040) V2 cos 37......(2)
(0.065)(5.5) = (0.065) (0.40863) V2 cos 65 + (0.040) V2 cos 37
0.3575=0.11216 V2 + 0.031945 V2
0.3575= 0.1441 V2
Solving for V2 (speed of the blue puck )
V2 =0.3575/0.1441 m/s
=2.4809 m/sec
Now substituting this value in eq (4) for speed of the red puck
V1 = 0.40863 V2 .......(4)
= 1.0137 m/s
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