Bouncy Collison (conserve P and KE) 1. V1 = 1.0 m/s mass = 1.00 kg and is headin

Question

Bouncy Collison (conserve P and KE)

1. V1 = 1.0 m/s mass = 1.00 kg and is heading to the right V2 = 0 m/s mass = 0.50 kg and no direction

After (solve for conserved P and KE)

V1 = ? mass = 1.0 kg mass = 1.00 kg direction = ? V2 = 4/3 mass = 0.50 kg direction = ?

2. V1 = 1.0 m/s mass = 0.50 kg and is heading to the right V2 = 0 m/s mass = 1.00 kg and no direction

After (solve for conserved P and KE)

V1 = ? mass = 0.50 kg direction = ? V2 = ? mass = 1.00 kg direction = ?

Sticky Collision (KE not conserved)

3. V1 = 1.0 m/s mass = 0.50 kg heading to right V2 = 0 m/s mass = 0.50 kg and has sticky attached to it

After (solve for conserved P and calculate delta KE)

Vboth = ? mass 0.50 kg + 0.50 kg they stick together

Explanation / Answer

Take the direction to the right to be positive.
a)m1*v1 + m2*v2 = 1m/s*1kg + 0*0.5kg = 1kg*m/s
1/2m1*v1^2 + 1/2m2*v2^2 = 1/2*1kg*(1m/s)^2 + 0 = 1/2J
Since the speed of v2 is 4/3 m/s, we get v1 = 1/3m/s, to the right. And v2 = 4/3 m/s, to the right.

b) m1*v1 + m2*v2 = 2m/s*0.5kg + 0*1kg = 1kg*m/s
1/2m1*v1^2 + 1/2m2*v2^2 = 1/2*0.5kg*(2m/s)^2 + 0 = 1J
By solving the two equations, we get v1 = -2/3 m/s, to the left, and v2 = 4/3 m/s, to the right.

c) (m1+m2)*v = 2m/s*0.5kg + 0*0.5kg = 1kg*m/s
v = 1kg*m/s / (0.5kg+0.5kg) = 1m/s, to the righ

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