A lighter block ( 1 kg ) and a heavier block ( 2 kg ) sit on a frictionless surf

Question

A lighter block (1 kg) and a heavier block (2 kg) sit on a frictionless surface. Both blocks are initially at rest. The same force of 4 N then pushes to the right on each block for a time of 3 s.

What are the initial momenta of the blocks?
p1i = 0  kg m/s
p2i = 0  kg m/s

What are the changes in momenta of the blocks?
p1 =  kg m/s
p2 =  kg m/s

What are the final momenta of the blocks?
p1f =  kg m/s
p2f =  kg m/s

What are the initial kinetic energies of the blocks?
K1i = 0  J
K2i = 0  J

What are the final kinetic energies of the blocks?
K1f =  J
K2f =  J

What are the changes in kinetic energy of the blocks?
K1i =  J
K2i =  J

How far does each block travel during the time of 3 s?
x1 =  m
x2 =  m

Explanation / Answer

A) Initial Momentum of both block s = 0

B) Net acceleration for lighter block = F/m = 4/1 = 4 m/s2

v= u + at = 0 + 4*3 = 12 m/s

del p1 = mv = 1*12 = 12 kg m/s

Net acceleration for lighter block = F/m = 4/2 = 2 m/s2

v= u + at = 0 + 2*3 = 6 m/s

del p1 = mv = 2*6 = 12 kg m/s

c) KE = 0.5mv^2 = 0

Inital KE for lighter and heavier mass = 0

d) Final KE of lighter Block = 0.5*1*12^2 = 72 J

Final KE of heavier block = 0.5*2*6^2 = 36 J

e) change in KE of lighter block = 72 - 0 = 72J

change in KE of heavier block = 36 - 0 = 36 J

f) Distance moved by lighter block

S = ut + 0.5at^2 = 0 + 0.5*4*3^2 = 18 m

Distance moved by Heavier block

S = 0 + 0.5*2*3^2 = 9 m

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