Question

The drawing shows a collision between two pucks on an air-hockey table. Puck A has a mass of 0.0240 kg and is moving along the x axis with a velocity of +6.27 m/s. It makes a collision with puck B, which has a mass of 0.0480 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. angles: a=65, b=37

Answer 1

We know that after the collision, Momentum will remain conserved, So

Using Momentum conservation in x-direction:

Pix = Pfx

m1*u1x + m2*u2x = m1*v1x + m2*v2x

m1 = mass of puck A = 0.024 kg = m

m2 = mass of puck B = 0.048 kg = 2*m

u1x = Initial velocity of m1 in x-direction = 6.27 m/sec

u2x = Initial velocity of m2 in x-direction = 0 m/sec

v1x = Final velocity of m1 in x-direction = v1*cos 65 deg

v2x = Final velocity of m2 in x-direction = v2*cos 37 deg

So,

m*6.27 + 2*m*0 = m*v1*cos 65 deg + 2*m*v2*cos 37 deg

6.27 = v1*0.423 + v2*1.597

Now Using Momentum conservation in y-direction:

Piy = Pfy

m1*u1y + m2*u2y = m1*v1y + m2*v2y

m1 = mass of puck A = 0.024 kg = m

m2 = mass of puck B = 0.048 kg = 2*m

u1y = Initial velocity of m1 in y-direction = 0 m/sec

u2y = Initial velocity of m2 in y-direction = 0 m/sec

v1y = Final velocity of m1 in y-direction = v1*sin 65 deg

v2y = Final velocity of m2 in y-direction = -v2*sin 37 deg

So,

m*0 + 2*m*0 = m*v1*sin 65 deg - 2*m*v2*sin 37 deg

0 = v1*0.906 - v2*1.204

v1 = v2*(1.204/0.906) = v2*1.329

Now Using both Bold equation

6.27 = v1*0.423 + v2*1.597

v1 = v2*1.329

Substitute value of v1 in 1st equation:

6.27 = v2*1.329*0.423 + v2*1.597

v2 = 6.27/[1.329*0.423 + 1.597]

v2 = 2.904 m/sec = final speed of puck B (Ans of Part (b))

Using this value

v1 = 2.904*1.329

v1 = 3.859 m/sec = final speed of puck A (Ans of part (a))

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