After being struck by a bowling ball, a 1.7 kg bowling pin sliding to the right at 3.8 m/s collides head-on with another 1.7 kg bowling pin initially at rest. Find the final velocity of the second pin in the following situations: a) The first pin moves to the right after the collision at 0.8 m/s. Answer in units of m/s.

3 m/s

Explanation:

Parameters given:

Mass of first bowling pin, m = 1.7 kg

Initial velocity of first bowling pin, u = 3.8 m/s

Final velocity of first bowling pin, v = 0.8 m/s

Mass of second bowling pin, M = 1.7 kg

Initial velocity of second bowling pin, U = 0 m/s

Let the final velocity of the second bowling pin be V

Using the principle of conservation of momentum:

Total initial momentum = Total final momentum

m*u + M*U = m*v + M*V

(1.7 * 3.8) + 0 = (1.7 * 0.8) + (1.7 * V)

6.46 = 1.36 + 1.7V

1.7V = 5.1

V = 5.1/1.7 = 3 m/s

one person pulls 90n one way and the other has 100n pulling the other way.

so the net force = 90 - 100 = -10n,

or the ans can be 10n if the picture shows -90n and 100n instead.