Question

A 2.0 kg block with a speed of 5.1 m/s collides with a 4.0 kg block that has a speed of 3.4 m/s in the same direction. After the collision, the 4.0 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 2.0 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 4.0 kg block ends up with a speed of 6.8 m/s. What then is the change in the total kinetic energy?

Answer 1

apply momentum conservation

Pi = P f

2*5.1 +4*3.4 = 2*V + 4*4.3

from here  V = 3.3 m/s

velocity of the 2.0 kg block immediately after the collision = 3.3 m/s

change in KE  

KEf - KEi

(1/2*2*3.3^2 +1/2*4*4.3^2 ) - ( 1/2*2*5.1^2 +1/2*4*3.4^2)

  = 1.26 energy  losss

otal kinetic energy change = 1.26 joule

(c) Suppose, instead, that the 4.0 kg b

2*5.1 +4*3.4 = 2*V + 4*6.8

V = -1.7 m/s

change in KE  

KEf - KEi

(1/2*2*1.7^2 +1/2*4*6.8^2 ) - ( 1/2*2*5.1^2 +1/2*4*3.4^2)

change in the total kinetic energy = 46.25 joule

( though it is not possible that energy after collision increasing )

let me know in a comment if there is any problem or doubts