A 7.6 kg block with a speed of 5.4 m/s collides with a 15.2 kg block that has a

Question

A 7.6 kg block with a speed of 5.4 m/s collides with a 15.2 kg block that has a speed of 3.6 m/s in the same direction. After the collision, the 15.2 kg block is observed to be traveling in the original direction with a speed of 4.5 m/s. (a) What is the velocity of the 7.6 kg block immediately after the collision? (b) By how much does the total kinetic energy of the system of two blocks change because of the collision? (c) Suppose, instead, that the 15.2 kg block ends up with a speed of 7.2 m/s. What then is the change in the total kinetic energy? (a) Number Units (b) Number Units (c) Number Units

Explanation / Answer

(a) By conservation of momentum,

m1u1+m2u2 = m1v1+m2v2

7.6*5.4+15.2*3.6 = 7.6*v1+15.2*4.5

v1 = (7.6*5.4+15.2*3.6-15.2*4.5)/7.6 = 3.6m/s

(b) The change in total K.E of system

change in KE = KE(f) - KE(i)

= (0.5m1v1^2+0.5m2v2^2) - (0.5m1u1^2+0.5m2v2^2)

= [0.5*7.6*(3.6)^2+0.5*15.2*(4.5)^2]-[0.5*7.6*(5.4)^2+0.5*15.2*(3.6)^2]

= - 6.156J

(c) By conservation of momentum

v1 =  (7.6*5.4+15.2*3.6-15.2*7.2)/7.6 = 1.8m/s

So, the new change in KE

[0.5*7.6*(1.8)^2+0.5*15.2*(7.2)^2] - [0.5*7.6*(5.4)^2+0.5*15.2*(3.6)^2] = 197J

The system loses energy in first collision and gains energy in srcond collision.

  

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