physics _ an inelastic collision on ice A hockey puck slides along the tee at a
ID: 1308757 • Letter: P
Question
physics _ an inelastic collision on ice
A hockey puck slides along the tee at a speed of 3.0 nVs to the RIGHT, striking a larger hockey puck which was not moving. It o a head on collision Assume there is no friction between the ice and the two masses After the collision, mass 1 is moving to the LEFT at 0 50 m/s while mass 2 is moving to the RIGHT at 1 75 m/s Define the +i direction as to the right Obtain the change in momentum of the mass 1. (It is a vector!) Obtain the change in momentum of the mass 2. (It s a vector!) Obtain the TOTAL momentum of the system before the collision (It's a vector!) Obtain the TOTAL momentum of the system after the collision. (It's a vector!) This is NOT a (perfectly) elastic collision. Obtain the energy lost during the collision.Explanation / Answer
Taking right as positive direction..
a.) Change= final -initial momentum (vectorized)
= (-5*0.5) - 5*3 = - 15 - 2.5 = -17.5 kgm/s
b.) Change = 10*1.75 -0 = 17.5 kgm/s
c.)Total momentum of system before collision = M(mass1) +M(mass2) = 5*3 + 0 = 15 kgm/s
d.)Total momentum of system after collision= M(mass1) +M(mass2) = -5*0.5 + 10*1.75 = 15 kgm/s
Hence, total momentum is conserved.
e.) initial energy = 1/2 (5* 3*3) = 22.5 kg (m/s)2
final energy = 1/2 (5* 0.5*0.5) + 1/2 (10 * 1.75*1.75) = 15.9375 kg (m/s)2
Hence, loss in energy =
22.5 kg (m/s)2 - 15.9375 kg (m/s)2 = 6.5625 kg (m/s)2
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