a block with mass m1=.500kg is released from rest on a frictionless track at a d

Question

a block with mass m1=.500kg is released from rest on a frictionless track at a distance h1=2.50m above the top of a table. It then collides elastically with an object having m2=1.00 kg that is initially at rest on the table. A) determine the velocities of the two objects after the collision. B) how high up the track does the .500kg object travel back after the collision? C) how far away from the bottom of the table does the 1.00kg object land, given that the height of the table is h2=2.00m? D) how far away from the bottom of the table does the .500kg object eventually land?

Explanation / Answer

a) Pe=Ke

m1gh=(1/2) m1 U1^2

U1=sqrt(2 gh)=7.0035m/s

Conservation of momentum

m1U1 +m2U2=m1V1 + m2V2

It helps to know that U2= 0

m1U1 = m1V1 + m2V2

What about conservation of kinetic energy? Ke=(1/2) mV^2

Ke1+Ke2=Ke1' + Ke2' again Ke2=0 and 1/2 term can be dropped

m1U1^2= m1V1^2 + m2V2^2

solving these equations we get

V1=(m1-m2)U1/(m1+m2) and since U1=sqrt(2 gh)= 7.0035 m/s

V1=(0.500 - 1.00) (7.0035)/(.500 +1.00)

V1=-2.33m/s

V2= 2m1U1/(m1+m2)

b) Again Ke(1/2)mV^2=mgh

h=(1/2)V^2 /g= 0.5 x (-2.33)^2 /9.81=.276 m

c) Time of fall =time of horizontal travel. Horizontal velocity is constant

t=sqrt(2 h2 / g)

t= sqrt( 2 x 2.0 / 9.81)=0.638 s

S= V2 t= 2m1U1/(m1+m2) t

S=[2 x 0.500 x 7.0035/(0.500 +1.00) ] 0.638

S=2.978m

d) The horizontal velocity will be the same in magnitude as it was after m1 colided with m2. The time of flight is the sam efor both of them so

S= V1t= 2.33 x 0.638= 1.486 m

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