A block with mass m 1 = 0.450 kg is released from rest on a frictionless track a

Question

A block with mass m1 = 0.450 kg is released from rest on a frictionless track at a distance h1 = 2.40 m above the top of a table. It then collides elastically with an object having mass m2 = 1.00 kg that is initially at rest on the table, as shown in the figure below.

v1 = m/s v2 = m/s A block with mass m1 = 0.450 kg is released from rest on a frictionless track at a distance h1 = 2.40 m above the top of a table. It then collides elastically with an object having mass m2 = 1.00 kg that is initially at rest on the table, as shown in the figure below. Determine the velocities of the two objects just after the collision. (Indicate the direction with the sign of your answer. Take the positive direction to be to the right.) How high up the track does the 0.450-kg object travel back after the collision? How far away from the bottom of the table does the 1.00-kg object land, given that the height of the table is h2 = 1.65 m? How far away from the bottom of the table does the 0.450-kg object eventually land?

Explanation / Answer

a) The velocity of m1 just before impact is found from

Pe=Ke

m1gh=(1/2) m1 U1^2

U1=sqrt(2 gh1)=6.85m/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)= 6.85 m/s

V1=(0.450 - 1.00) ( 6.85)/(0.450 +1.00)

V1=-2.59 m/s

V2= 2m1U1/(m1+m2) = 4.25 m/s

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

h=(1/2)V^2 /g= 0.5 x (2.59)^2 /9.81=.342 m

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

t=sqrt(2 h2 / g)

t= sqrt( 2 x 1.65 / 9.81)=0.579 s = 0.58

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

S=[2 x 0.450 x 6.85/(0.45 +1.00) ] 0.58 =

S=4.25m

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= 2.59 x 0.58= 1.50 m

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