A block with a mass of m 1 = 5.6 kg is pushed to the right with a force of 103 N

Question

A block with a mass of m1 = 5.6 kg is pushed to the right with a force of 103 N for a distance of 7.9 m across a horizontal frictionless surface, after which the force is removed. The 5.6-kg block then collides with a second block with a mass m2= 1 kg, which is initially at rest. The two blocks stick together after the collision.

1)

Determine the change in the internal energy of the system of blocks during the collision.

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2)

Now assume that the 1-kg block is traveling at 2 m/s to the left before the collision. The two blocks will still stick together after the collision. Determine the change in the internal energy of the system of blocks.

Explanation / Answer

1) Energy transferred to the block m1= work done by the external force= 103*7.9= 813.7 Nm

so, K.E= 0.5mv^2=813.7

v^2=813.7/(5.6*.5)= 290.6 m/s

v=17.047 m/s

So,mech energy of block m1 before collision=(P.E+ K.E) of m1= 813.7 Nm

mech energy of block m2 before collision=(P.E+ K.E) of m2= 0

Now, by momentum coservation

m1v1+m2v2=MV

5.6*17.047+0=6.6*V

V=14.464 m/s

mech energy of the system after collision= (P.E+ K.E)= 0+.5*6.6*14.464^2=690.38 Nm

Change in the internal energy of the system after collision= initial energy-final energy

   = 813.7-690.38=123.316 Nm

2) now,

Now, by momentum coservation

m1v1+m2v2=MV

5.6*17.047+1*(-2)=6.6*V

V=14.16 m/s

mech energy of block m1 before collision=(P.E+ K.E) of m1= 813.7 Nm

mech energy of block m2 before collision=(P.E+ K.E) of m2= 0+.5*1*(-2)^2=2Nm

Change in the internal energy of the system after collision= final energy- initial energy

   = .5*6.6*14.16^2-813.7-2= 153.93 Nm

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