The drawing shows a collision between two pucks on an air-hockey table. Puck A h

Question

The drawing shows a collision between two pucks on an air-hockey table. Puck A has a mass of 0.020 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 0.055 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.

(a) Find the final speed of puck A.
m/s

(b) Find the final speed of puck B.
  m/s

The drawing shows a collision between two pucks on an air-hockey table. Puck A has a mass of 0.020 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 0.055 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. (a) Find the final speed of puck A. m/s (b) Find the final speed of puck B. m/s

Explanation / Answer

Given that,

m1 = 0.020 kg and u1 =5.5 m/s

m2 = 0.055 kg and u2 = 0 (its in rest position)

Momentum along x axis before collision = m1u1 = 0.020 x 5.5 = 0.11 and along Y is 0 as u2 =0

Let V1 and V2 be the velocities of puck A and B respectively, aft collision occured.

Momentum conservation along Y axis:

m1V1 cos (90 - 65) = m2 V2 Cos (90 - 37)

(0.020) V1 Cos (25) = (0.055) V2 Cos (53)

V1 = (0.055 x 0.6 / 0.020 x 0.9)V2 gives us

V1 = 1.83V2

Momentum conservation along X axis

m1u1 = m1 V1 Cos (65) + m2 V2 Cos (37)

0.020x5.5 = 0.020 x V1 x (0.422) + 0.055 x V2 x (0.79)

0.11 = 0.00844 V1 + 0.004345 V2

We have V1 = 1.83V2

0.11 = 0.00844 (1.83V2) + 0.004345 V2 = 0.019 V2

V2 = 5.78 m/s

V1 = 1.83(5.78) = 10.577 m/s

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