Question

a 0.6 kg ball traveling horizontally on a frictionless surface approaches a wall at a speed of 20m/s perpendicularly to the wall and rebounds with 70% of its initial kinetic energy. What is the magnitude of the impulse exerted on the ball by the wall?

A. 12.0 kg*m/s

B. 1.96 kg*m/s

C. 22.0 kg*m/s

D. 20.4 kg*m/s

E. 3.60 kg*m/s

Answer 1

According to the first law of Thermodynamics : Energy can be neither created nor destroyed. It can be converted from one form to other form. Potential Energy and Kinetic Energy are the two basic forms.

Balancing the Energies Potential Energy and Kinetic Energy.

Mass of the Ball = 0.6Kg.

Speed of the ball = 20 m/s.

The surface is frictionless.

Kinetic Energy for the rebounds = 70 % of initial Kinetic Energy

0.5 * 0.6 * v² = 70 * 0.5 * 0.6 * 20²

v² = 70 * 20²

^{v =}

⁼

= 16.7332 m/s

The magnitude of the impulse exerted on the ball by the wall at a defined interval of time with the momentum of object P is impulse J.

so J = F(t) dt = P

Because the ball rebounds elastically, we know that kinetic energy is also conserved in the collision. Given the ball rebounds, it has 70% conserved kinetic energy but velocity moves in the opposite direction.

so P = mvf - mvi

            = 0.6 *-16.7332 - 0.6 * 20

            = 0.6 (-16.7332-20)

           = 0.6 * - 36.7332

           = -22.04 Kg m/s.

           Note that the negative sign indicates direction. So we have:

            J=?p

          ?=-22.04 kgm/s

Answer is C.