Collision finding final velocity

  • Thread starter mr1709
  • Start date
  • #1
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1. A baseball of mass 0.30kg is pitched at 70m/s (fwd) at a batter. The ball knocks the stationary 1.7kg ball out of the batters hands and the ball rebounds at 48m/s (backwards). What is the final velocity of the bat as it leaves the batters hand?

variables
m1 = 0.30 vi= 70 vf = -48

m2= 1.7 vi = 0 vf = ?


3. This issue i ran into in this problem is i get two different answers depending on the method used to solve.

Method one:

m1vi1 = m1vf1 + m2vf2
0.30(70) = 0.30(-48) + 1.7(vf)
vf = 20.823

Method two: Using a shortcut formula derived in textbook

for reference, this is the formula im talking about that the book derived

http://hyperphysics.phy-astr.gsu.edu/hbase/imgmec/elacol18.gif


formula : vf2= (m2-m1)/(m2+m1) * vi2 + (2m1)/(m1 + m2) * vi1
first part of formula cancels due to vi2 being 0. Therefore:
vf2= 0.60/2 * 70
= 21 m/s

How come the answers differ? I dont have the correct answer as it wasnt given in the worksheet so im not sure which method is correct
 

Answers and Replies

  • #2
35,628
12,168
Your second formula assumes energy is conserved in the collision. It doesn't have to be.
 

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