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 0.0480 kg and is moving along the x axis with a velocity of +6.98 m/s. It makes a collision with puck B, which has a mass of 0.0960 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. Find the speed of (a) puck A and (b) puck B.

The drawing shows a collision between two pucks on an air-hockey table. Puck A has a mass of 0.0480 kg and is moving along the x axis with a velocity of +6.98 m/s. It makes a collision with puck B, which has a mass of 0.0960 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. Find the speed of (a) puck A and (b) puck B. 65 37 At Before collision After collision (a) Number Units (b) Number Units

Explanation / Answer

using momentum conservation

in X-direction

Pxi = Pxf

mA = m & mb = 2m

m*Vaxi + 2m*Vbxi = m*Vaf*cos 65 deg + 2m*Vbf*cos 37 deg

Vaxi + 2*Vbxi = Vaf*cos 65 deg + 2*Vbf*cos 37 deg

Vbxi = 0

6.98 + 0 = Vaf*0.423 + Vbf*1.597

in y-direction

Pyi = Pyf

mA = m & mb = 2m

m*Vayi + 2m*Vbyi = 2m*Vaf*sin 65 deg - 2*m*Vbf*sin 37 deg

Vayi + 2*Vbyi = Vaf*sin 65 deg - 2*Vbf*sin 37 deg

Vbyi = Vayi = 0

0 + 0 = [Vaf*0.906 - Vbf*1.204]

Vaf = Vbf*(1.204/0.906) = 1.329*Vbf

using bolded equation

6.98 + 0 = Vaf*0.423 + Vbf*1.597

6.98 + 0 = Vbf*1.329*0.423 + Vbf*1.597

Vbf = 6.98/(1.329*0.423 + 1.597) m/sec = 3.233 m/sec

Vaf = 1.329*Vbf = 1.329*3.233 m/sec = 4.297 m/sec

Puck A = 4.297 m/sec

Puck B = 3.233 m/sec

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