not really sure how to go about this problem: Problem: a block of unknown mass m

Question

not really sure how to go about this problem:
Problem: a block of unknown mass m2 is at rest on a long frictionless table. A block of mass m1 =.5kg is projected toward the first block, with speed v1 = 2.0 m/s. Assume all collisions are perfectly elastic.
A) find the value of m2 for which both blocks move with the same speed but opposite directions after the collision.
B) find the speed of either block after the collision
C) find the fraction of the original K.E. that is transferred from mass m1 to mass m2.
D)After some time, the blocks both bounce off of perfect springs, and so they both change directions but keep their same speeds as before they hit the springs. Thus the blocks are now moving towards each other. What will their respective velocities be after the next ensuing collision? ------I think this is V =0, but why?

Explanation / Answer

we have that equations of velocities of blocks after collision. v1=(u1(m1-m2)+2u2m2)/(m1+m2). v2=(u2(m2-m1)+2u1m1)/(m1+m2). -----------a a) so we have v1=-v2. so that u1(m1-m2)+2u2m2=-u2(m2-m1)-2u1m1. where u1=2.0 and u2=0. so 2(m1-m2)=-2*2*m1. so that m1-m2=-2m1. so 3m1=m2. so m2=1.5kg. b) it is v2=2*2*0.5/(1.5+0.5)=1(m/s) so v1=-1(m/s). c)H=K2/(K1+K2)=(m2v2^2)/(m1v1^2+m2v2^2)=m2/(m1+m2)=0.75. d)its another elastic collision. where u1=1m/s and u2=-1m/s. plug in we have. v1=(-1*1-2*1*1.5)/2=-2(m/s). v2=(-1*1+2*1*0.5)/2=0(m/s)

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