You toss a 0.40-kg ball at 6.0 m/s to a 17-kg dog standing on an iced-over pond.

Question

You toss a 0.40-kg ball at 6.0 m/s to a 17-kg dog standing on an iced-over pond. The dog catches the ball and begins to slide on the ice. Call the direction in which you throw the ball the +x direction.

Measured from the Earth reference frame, how much of the original kinetic energy of the system is convertible?

Measured from an inertial reference frame in which the ball's kinetic energy does not change, how much of the original kinetic energy of the system is convertible?

Convertible Energy is the portion of the kenetic energy that can be converted to internal energy.

Explanation / Answer

apply the law of conservation of momentum as

m1u1 +m2u2 = m1v1 + m2V2

0.4* 6 = (17 + 0.4) * V2

V2 = 0.138 m/s

----------------------------
initial KE of the ball = 0.5 * 0.4 * (6-v)^2

INitial KE 0.5 * 0.4* 6^2

final KE of ball = 0.5 * 0.4* (0.138-v)^2

so

6-V = 0.138 -V

2V = 6.138

V = 3.07 m/s

convertable energy = 0.5* 0.4* (6)^2 - 0.5 * 0.4* 3.07^2

= 5.31 Joules

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