A 7.3 kg block with a speed of 5.1 m/s collides with a 14.6 kg block that has a

Question

A 7.3 kg block with a speed of 5.1 m/s collides with a 14.6 kg block that has a speed of 3.4 m/s in the same direction. After the collision, the 14.6 kg block is observed to be traveling in the original direction with a speed of 4.3 m/s. (a) What is the velocity of the 7.3 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 14.6 kg block ends up with a speed of 6.8 m/s. What then is the change in the total kinetic energy?

Explanation / Answer

(a) This is an elastic collision, and from the law of conservation of momentum, the speed of the 19kg block is:

mv(i) + mv= mv(f) + mv(f)

Solved for v(f):

v(f) = [mv(i) + mv - mv(f) ]/ m
= [(4.5kg)(11m/s) + (19kg)(6.8m/s) - (4.5kg)(3.7m/s)] / 19kg
= 8.5m/s

(b) The change in the total kinetic energy of the system is:

KE = [KE+ KE](f) - [KE+ KE](i)
= [0.5mv² + 0.5mv²](f) - [0.5mv² + 0.5mv²](i)
= [0.5(4.5kg)(3.7m/s)²+ 0.5(19kg)(8.5m/s)²] - [0.5(4.5kg)(11m/s)² + 0.5(19kg)(6.8m/s)²]
= -6.424J

(c) If m2ends up with a speed of 6.8m/s (I guess this changes the speed of mas well. So m1's new speed woud be:

v1(f) = [mv(i) + mv - mv(f) ]/ m
= [(7.3kg)(5.1m/s) + (14.6kg)(6.8m/s) - (7.3kg)(4.3m/s)] / 14.6kg
= 7.2m/s

So the new change in KE is:

KE = [KE+ KE](f) - [KE+ KE](i)
= [0.5mv² + 0.5mv²](f) - [0.5mv² + 0.5mv²](i)
= [0.5(7.3kg)(7.2m/s)²+ 0.5(14.6kg)(6.8.m/s)²] - [0.5(7.3kg)(5.1m/s)² + 0.5(14.6kg)(3.4m/s)²]
= 347.44J

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