A m1 = 0.500 kg block is released from rest at the top of a frictionless track h

Question

A m1 = 0.500 kg block is released from rest at the top of a frictionless track h1 = 3.10 m above the top of a table. It then collides elastically with a 1.00 kg block that is initially at rest on the table. (a) Determine the velocities of the two objects just after the collision. velocity of m1 ? m/s velocity of m2 ? m/s (b) How high up the track does the 0.500 kg object travel back after the collision? ?m (c) How far away from the bottom of the table does the 1.00 kg object land, given that the table is 1.90 m high? ? m (d) How far away from the bottom of the table does the 0.500 kg object eventually land? ?m

Explanation / Answer

a) The velocity of m1 just before impact is found from
Pe=Ke
m1gh=(1/2) m1 U1^2
U1=sqrt(2 gh)=sqrt(2 *9.8*3.1)=7.8 m/s

Conservation of momentum
m1U1 +m2U2=m1V1 + m2V2
It helps to know that U2= 0

m1U1 = m1V1 + m2V2
also
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 (See ref 1)
solving these equations we get

V1=(m1-m2)U1/(m1+m2) and since U1=sqrt(2 gh)= 7.8 m/s
V1=(0.50 - 1.00) (7.8)/(0.5 +1.00)
V1=-2.6m/s--> Answer

V2= 2m1U1/(m1+m2)
V2 = (0.50 *2) (7.8)/(0.5 +1.00)
V2 = 5.2 m/s--> Answer

b) Again Ke(1/2)mV^2=mgh
h=(1/2)V^2 /g= 0.5 x (2.6)^2 /9.81=0.344 m--> Answer

c) Time of fall =time of horizontal travel. Horizontal velocity is constant
t=sqrt(2 h2 / g)
t= sqrt( 2 x 1.9 / 9.81)=0.62 s
S= V2 t= 2m1U1/(m1+m2) t=
S=[2 x 0.50 x 7.8/(0.5 +1.00) ] 0.62
S=3.24m--> Answer

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

S= V1t= 2.6 x 0.62= 1.612 m--> Answer

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