Question

A 2.7 kg block with a speed of 5.4 m/s collides with a 5.4 kg block that has a speed of 3.6 m/s in the same direction. After the collision, the 5.4 kg block is observed to be traveling in the original direction with a speed of 4.5 m/s. (a) What is the velocity of the 2.7 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 5.4 kg block ends up with a speed of 7.2 m/s. What then is the change in the total kinetic energy?

Part a: 3.6 m/s

Part b: -2.19 J

I can not figure out part c.

Answer 1

here,

mass of block 1 , m1 = 2.7 kg

mass of block 2 , m2 = 5.4 kg

initial speed of block 1 , u1 = 5.4 m/s

initial speed of block 2 , u2 = 3.6 m/s

final speed of second block , v2 = 4.5 m/s

a)

let the final velocity of 2.7 kg object be v1

using conservation of momentum

m1 * u1 + m2 * u2 = m1 * v1 + m2 * v2

2.7 * 5.4 + 5.4 * 3.6 = 2.7 * v1 + 5.4 * 4.5

solving for v1

v1 = 3.6 m/s

the final velocity of 2.7 kg is 3.6 m/s

b)

the change in total energy , dKE = KEf - KEi

dKE = (0.5 * m1 * v1^2 + 0.5 * m2 * v2^2) - (0.5 * m1 * u1^2 + 0.5 * m2 * v2^2)

dKE = (0.5 * 2.7 * 3.6^2 + 0.5 * 5.4 * 4.5^2) - ( 0.5 * 2.7 * 5.4^2 + 0.5 * 5.4 * 3.6^2)

dKE = - 2.19 J

c)

when v2 = 7.2 m/s

let the final velocity of 2.7 kg object be v1

using conservation of momentum

m1 * u1 + m2 * u2 = m1 * v1 + m2 * v2

2.7 * 5.4 + 5.4 * 3.6 = 2.7 * v1 + 5.4 * 7.2

solving for v1

v1 = - 1.8 m/s

the final velocity of 2.7 kg is - 1.8 m/s

the change in total energy , dKE = KEf - KEi

dKE = (0.5 * m1 * v1^2 + 0.5 * m2 * v2^2) - (0.5 * m1 * u1^2 + 0.5 * m2 * v2^2)

dKE = (0.5 * 2.7 * 3.6^2 + 0.5 * 5.4 * 4.5^2) - ( 0.5 * 2.7 * 1.8^2 + 0.5 * 5.4 * 7.2^2)

dKE = - 72.1 J