On a frictionless surface, a 1\" chunk of play-doe of mass 7.0 kg, is traveling

Question

On a frictionless surface, a 1" chunk of play-doe of mass 7.0 kg, is traveling to the right at 6.0 m/s and smashes into a 2nd chunk of play-doe of mass 18.0 kg which is traveling to the left at 4.0 m/s. Immediately after the collision, the 1st and 2nd chunks of play-doe move off as one big mass of play-doe at some velocity. a) After the collision, what is the velocity of the big mass made up of the 1 and 2nd chunks of play-doe, indicate magnitude and direction left or right? b) Calculate the total kinetic energy of the 1st and 2nd chunks of play-doe before the collision. C) Calculate the kinetic energy of the 1" and 2nd chunks of play-doe as one big chunk after the collision. d) was this collision elastic or inelastic; explain your answer by using your results for the kinetic energy before and after the collision. 2. Now instead the two masses are made of wood resulting in a change in the collision process as follows: On a frictionless surface, a 1 wooden ball of mass 7.0 kg, is traveling to the right at 6.0 m/s and collides with a 2nd wooden ball of mass 18.0 kg which is traveling to the left at 4.0 m/s. Immediately after the collision, the 18 wooden ball reverses direction and moves left at a velocity of 3.0 m/s a) After the collision, what is the velocity of the 2nd wooden ballindicate magnitude and direction left or right? b) Calculate the total kinetic energy of the 1st and 2nd wooden balls before the collision. C) Calculate the kinetic energy of the 1" and 2nd wooden balls after the collision d) Was this collision elastic or inelastic; explain your answer by using your results for the kinetic energy before and after the collision

Explanation / Answer

1.

a)

By conservation of momentum

M1V1+M2V2=(M1+M2)V

7*6+18*(-4)=(7+18)*V

V=-1.2 m/s

V=1.2 m/s (travelling to the left)

b)

Initial Kinetic energy before collision

Kinitial=(1/2)M1V12+(1/2)M2V22=(1/2)*7*62+(1/2)*18*(-4)2

Kinitial =270 J

c)

final kinetic energy after collision

Kfinal= (1/2)(M1+M2)V2=(1/2)*(7+18)*1.22

Kfinal=18 J

d)

The collision is inelastic ,since Kinitial not equal to Kfinal i,e converted to some other form of energy.

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