Question

A 7.2 kg block with a speed of 10 m/s collides with a 19 kg block that has a speed of 5.4 m/s in the same direction. After the collision, the 19 kg block is observed to be traveling in the original direction with a speed of 5.4 m/s. (a) What is the velocity of the 7.2 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 19 kg block ends up with a speed of 4.7 m/s. What then is the change in the total kinetic energy?

Answer 1

Where m₁ = Mass of first body ,=7.2kg.
m₂ = Mass of second body,=19kg.
u₁ = Initial velocity of first body,=10m/s
u₂ = Initial velocity of second body,=5.4m/s
v₁ = Final velocity of first body,=?
v₂ = Final velocity of second body=5.4m/s.

it is a elastic collision.

so,m₁u₁+m₂u₂=m₁v₁+m₂v₂.

(7.2)(10)+(19)(5.4)=(7.2)(v₁)+(19)(5.4).

174.6=7.2v₁+102.6.

   7.2v₁=174.6-102.6

     :.v₁=72/7.2

    :.v₁=10m/s.

2.change in kinetic energy is given by=m₁v₁²/2+m₂v₂²/2.

                                                         =[(7.2)(10²)/2]+[(19)(5.4²)/2].

                                     :.K.E.=360+277.02

                                    :.K.E=637.02J.

3.change in kinetic energy is=m₁v₁²/2+m₂v₂²/2.

                                           =[(7.2)(10²)/2]+[(19)(4.7²)/2].

                                :.K.E=360+209.855

                             :.K.E.=569.855J.