A block with mass m1 = 0.500 kg is released from rest on a frictionless track at

Question

A block with mass m1 = 0.500 kg is released from rest on a frictionless track at a distance h1 = 2.50 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. a. Describe the energy and momentum transformations that occur throughout the motion of both blocks. Be thorough in your explanations for full credit. b. 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 = 2.00 m? c. How far away from the bottom of the table does the 0.500-kg object eventually land?

Explanation / Answer

here,

M1 = 0.5 kg
M2 = 1 kg
H1 = 2.5 m
H2 = 2 m
V1 = 0 m/s
V2 = 0 m/s

A:)

From Conservation Energy

Potential Energy at H = Kinectic energy at bottom

M1*g*H1 = 0.5 * M1 * V1'^2

V1 = Sqrt( 2*g*h1 )
V1 = Sqrt( 2*9.8*2.5 )
V1 = 7 m/s

From Conservation of Momentum:
Momentum Before Collision = Momentum after collision

M1V1 + M2V2 = M1V1' + M2V2'
M1V1 = M1V1' + M2V2'

also from Conservation of Energy we have :

0.5M1V1^2 + 0.5 M2V2^2 = 0.5M1V1'^2 + 0.5 M2V2'^2
M1V1^2 = M1V1'^2 + M2V2'^2

on subtituting we get :

V1' = (m1-m2)*V1/(m1+m2)
V1' = -0.5*7 / (1.5)
V1' = -2.33 m/s

also,

V2' = 2m1*V1/(m1+m2)
V2' = 2*0.5*7 / (1.5)
V2' = 4.66 m/s

B)
Time taken ball to meet the ground from H2

T = sqrt(2h/g)
T = sqrt(2*2/9.8)
T = 0.63 s

Distance X = V2' * t = 4.66 * 0.63 = 2.935 m

the distanc travelled by block 2 when fallen from table is 2.025 m

C)

Similarly
X = V1' * t = 2.33 * 0 .63 = 1.46 m

the distanc travelled by block 2 when fallen from table is 1.46 m

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