A 2.9 kg block with a speed of 13 m/s collides with a 15 kg block that has a spe

Question

A 2.9 kg block with a speed of 13 m/s collides with a 15 kg block that has a speed of 5.7 m/s in the same direction. After the collision, the 2.9 kg block is observed to be traveling in the original direction with a speed of 5.8 m/s. (a) What is the velocity of the 15 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 2.9 kg block ends up with a speed of 4.9 m/s. What then is the change in the total kinetic energy?

Explanation / Answer

first apply momentum conservation

Pi = Pf

take first block velocity direction as positive so   

2.9* 13 +15*5.7 = 2.9* 5.8 + 15 *V

V = 7.092 it comes out positve so it is also moving in same direction as 2.9 kg

change in KE

KEf - KEi

( 0.5 *2.9*5.82  + 0.5 * 15 *7.0922 ) - (0.5*2.9*132 + 0.5*15*5.72)

= -62.72352 J

negetive   means energy lost in the collision

apply again momentum conservation

2.9* 13 +15*5.7 = 2.9* 4.9 + 15 *V

V = 7.226

change in KE

KEf - KEi

( 0.5 *2.9*4.92  + 0.5 * 15 *7.226 2 ) - (0.5*2.9*132 + 0.5*15*5.72)

= -62.29743 J

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