Question

1. a curling stone with a mass of 4.5 kg is pushed down a sheet of frictionless ice at a velocity of 4.0 m/s. it collides with a second curling stone of mass 8.5 kg. after the collision, the first stone slides backwards with a velocity of -0.5 m/s.

a) draw a picture showing the situation described before and after the collision. draw arrows showing the velocities of each cart.
b) calculate the total momentum of the system before the collision.
c) calculate the total momentum of the system after the collision.
d) calculate the velocity of the second stone after the collision.

Explanation:

Step1: Determine the formula for momentum

The formula for momentum \(p = mv\), where \(m\) is mass and \(v\) is velocity.

Step2: Calculate the total momentum before the collision

For the first stone, \(m_1 = 4.5\space kg\) and \(v_1=4.0\space m/s\), so \(p_1=m_1v_1 = 4.5\times4.0=18\space kg\cdot m/s\). The second stone is initially at rest, so \(v_2 = 0\space m/s\) and \(p_2=m_2v_2=8.5\times0 = 0\space kg\cdot m/s\). The total momentum before collision \(P_{before}=p_1 + p_2=18+0 = 18\space kg\cdot m/s\).

Step3: Calculate the total momentum after the collision

For the first stone after collision, \(m_1 = 4.5\space kg\) and \(v_1'=- 0.5\space m/s\), so \(p_1'=m_1v_1'=4.5\times(-0.5)=-2.25\space kg\cdot m/s\). Let the velocity of the second stone after collision be \(v_2'\). Then \(p_2'=m_2v_2'=8.5v_2'\). By conservation of momentum \(P_{before}=P_{after}\), so \(18=-2.25 + 8.5v_2'\).

Step4: Solve for \(v_2'\)

Rearrange the equation \(18+2.25=8.5v_2'\), so \(v_2'=\frac{18 + 2.25}{8.5}=\frac{20.25}{8.5}\approx2.38\space m/s\)

Answer:

b) The total momentum of the system before the collision is \(18\space kg\cdot m/s\).
c) The total momentum of the system after the collision is \(18\space kg\cdot m/s\) (by conservation of momentum).
d) The velocity of the second stone after the collision is approximately \(2.38\space m/s\)