Please be detailed Bouncy Collision a. Before: V1=1.0m/s 1.00kg and is heading t
ID: 2023108 • Letter: P
Question
Please be detailed
Bouncy Collision
a. Before: V1=1.0m/s 1.00kg and is heading to the rightV2=0m/s 0.5kg no direction
After(Solve for conserved P and KE): V1=? (1.00kg) direction? V2=4/3 ( 0.50kg) in what directon?
b. Before: V1=2.0m/s 0.50kg heading to the rightV2=0m/s 1.00kg no direction
After(solve for conserved P and KE): V1=? (0.50kg) what directon?V2=? (1.00kg) what direction?
STICKY COLLISION (KE not conserved)
c. V1=2.0m/s 0.50kg HEADING TO THE RIGHTV2=0m/s 0.50kg has the sticky attached to it
Vboth=? (0.50kg+0.50kg)they stick to each other
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 right.