A 2.5 kg block with a speed of 14 m/s collides with a 20 kg block that has a spe
ID: 1465433 • Letter: A
Question
A 2.5 kg block with a speed of 14 m/s collides with a 20 kg block that has a speed of 3.9 m/s in the same direction. After the collision, the 2.5 kg block is observed to be traveling in the original direction with a speed of 8.2 m/s. (a) What is the velocity of the 20 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 2.5 kg block ends up with a speed of 6.9 m/s. What then is the change in the total kinetic energy?
Explanation / Answer
from the conservation of momentum
m1u1+ m2 u2= m1v1 + m2 v2
2.5( 14) + 20( 3.9) = 2.5(8.2) + 20 kg v2
v2 = 4.62 m/s
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initial kinetic energy
1/2 m1 u1^2 + 1/2 m2u^2 = 1/2( 2.5) ( 14)^2 + 1/2 ( 20 )( 3.9)^2 = 397.1 J
final kinetic energy
1/2 m1 v1^2 + 1/2 m2v^2 = 1/2( 2.5) ( 8.2)^2 + 1/2 ( 20 )( 4.62)^2 = 297.49 J
change in kientic energy
del KE = 397.1 J-297.49 J = 99.61 J
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m1u1+ m2 u2= m1v1 + m2 v2
2.5( 14) + 20( 3.9) = 2.5(6.9) + 20 kg v2
v2 = 4.78 m/s
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initial kinetic energy
1/2 m1 u1^2 + 1/2 m2u^2 = 1/2( 2.5) ( 14)^2 + 1/2 ( 20 )( 3.9)^2 = 397.1 J
final kinetic energy
1/2 m1 v1^2 + 1/2 m2v^2 = 1/2( 2.5) ( 8.2)^2 + 1/2 ( 20 )( 4.78)^2 = 313.25J
change in kientic energy
del KE = 397.1 J-313.25 J = 83.84 J
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