amountLost= initial KE-final KE
= 1/2 *3*5^2+1/2 *2*2^2+1/2 *3*1^2=43Joules
check that.
A 3kg mass sliding on a frictionless surface has a velocity of 5m/s east when it undergoes a one-dimensional inelastic collision with a 2kg mass that has an initial velocity of 2m/s west. After the collision the 3kg mass has a velocity of 1m/s east. How much kinetic energy does the two-mass system lose during the collision?
2 answers
Given:
M1 = 3kg, V1 = 5m/s.
M2 = 2kg, V2 = -2m/s.
V3 = Velocity of M1 after collision.
V4 = 1m/s = Velocity of M2 after collision.
Momentum before = Momentum after.
M1*V1 + M2*V2 = M1*V3 + M2*1.
3*5 + 2*(-2) = 3*V3 + 2*1,
V3 = 3 m/s.
KEb = 0.5*3*5^2 - 0.5*2*2^2 . = KE before collision.
KEa = 0.5*3*3^2 + 0.5*2*1^2. = KE after collision.
KE lost = KEb - KEa.
k
ea
M1 = 3kg, V1 = 5m/s.
M2 = 2kg, V2 = -2m/s.
V3 = Velocity of M1 after collision.
V4 = 1m/s = Velocity of M2 after collision.
Momentum before = Momentum after.
M1*V1 + M2*V2 = M1*V3 + M2*1.
3*5 + 2*(-2) = 3*V3 + 2*1,
V3 = 3 m/s.
KEb = 0.5*3*5^2 - 0.5*2*2^2 . = KE before collision.
KEa = 0.5*3*3^2 + 0.5*2*1^2. = KE after collision.
KE lost = KEb - KEa.
k
ea
1 answer