The conservation of momentum P states that the amount of momentum remains constant when there are not external forces.
We don't have external forces, so:
[tex]P_0 = P_1\\m_bv_{0b}+m_pv_{0p}=m_bv_{1b}+m_pv_{1p}\\[/tex]
Where:
Solving for v0b:
[tex]v_{0b} =\dfrac{m_bv_{1b}+m_pv_{1p}- m_pv_{0p}}{m_{b}}\\\\v_{0b} =\dfrac{(7\;kg)(4\;m/s)+(2\;kg)(6\;m/s)- (2\;kg)(0 \;m/s)}{7\;kg}\\v_{0b}=\dfrac{40}{7}\;m/s\\\\\boxed{v_{0b}\approx5.71\;m/s}[/tex]
The original velocity of the ball is 5.71 m/s.
The principle of conservation of momentum: In a closed system, The total momentum before collision is equal to total momentum after collision.
From the principle of conservation of momentum,
MU+mu = MV+mv.................... Equation 1
Where M = mass of the bowling ball, m = mass of the pin, U = initial velocity of the bowling ball, u = initial velocity of the pin, V = final velocity of the bowling ball, v = final velocity of the pin.
From the question,
Given: M = 7 kg, m = 2 kg, u = 0 m/s (at rest), v = 6.0 m/s, V = 4 m/s.
Substitute these values into equation 1 and solve for U
7(U)+2(0) = 7(4)+2(6)
7U = 28+12
7U = 40
U = 40/7
U = 5.71 m/s.
Hence, The original velocity of the ball is 5.71 m/s.
Learn more about velocity here: https://brainly.com/question/6237128
