A 7.9 kg block with a speed of 15 m/s collides with a 16 kg block that has a spe

Question

A 7.9 kg block with a speed of 15 m/s collides with a 16 kg block that has a speed of 4.4 m/s in the same direction. After the collision, the 16 kg block is observed to be traveling in the original direction with a speed of 10 m/s. (a) What is the velocity of the 7.9 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 16 kg block ends up with a speed of 4.9 m/s. What then is the change in the total kinetic energy?

Explanation / Answer

applying the conservation of momentum

m1u1 + m2u2 =m1v1 +m2+v2

7.9 *15 +16 *4.4 = 7.9*10 + 16* v2

v2 = 6.86 m/s

change in kinetic energy = initial kinetic energy of the block 1 - final kinetic energy of block 1

= 0 .5* m1*v1^2 - .5 *m1 *u1^2 = 493.75

change in kinetic energy of block 2 - final kinetic energy of block 2

= 0.5*16*4.4^2 - 0.5*16*6.86^2 = -221.5

if the block ends up with 4.9 m/s = -37.2

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