I'm having a hard time doing this problem for Physics

Question

A 6.0 kg bowling ball moving at 3.5 m/s to the right makes a collision, head-on, with a sta-tionary 0.70 kg bowling pin. If the ball is moving 2.77 m/s to the right after the collision, what will be the velocity (magnitude and direction) of the pin?

Answer 1

This seems like it should be a two dimensional problem because you envision the ball and pin having the ability to move in any direction, however, all motion is occurring to the right therefore, it is really a 1 dimensional problem - there is no motion in any direction but to the right.

The momentum before the collision is given by P₁ = (m_ball)(v_ball) + (m_pin)(v_pin) = (6.0)(3.5) + (0.70)(0) = 21 kgm/s and the momentum immediately after the collision is given by P2 = (m_ball)(v_ball) + (m_pin)(v_pin) = (6.0)(2.77) + (0.70(v_pin).

Due to the conservation of momentum, P₁ = P₂ so 21 = (6.0)(2.77) + (0.70(v_pin).

21 = 16.62 +0.70v_pin

4.38 = 0.70v_pin

v_pin = 6.3 m/s to the right