Question

question 15
a 7.30 - kg bowling ball strikes a 1.60 - kg pin at rest head - on. before the collision, the velocity of the ball is + 6.00 m/s. after the collision, the velocity of the ball is + 5.40 m/s. what is the velocity of the pin after the collision?
+ 1.2 m/s
+ 2.7 m/s
+ 3.2 m/s
+ 0.6 m/s
+ 5.4 m/s

Explanation:

Step1: Apply the law of conservation of momentum

The law of conservation of momentum states that \(m_1u_1 + m_2u_2=m_1v_1 + m_2v_2\). Here, \(m_1 = 7.30\space kg\) (mass of the bowling ball), \(u_1=+ 6.00\space m/s\) (initial velocity of the ball), \(m_2 = 1.60\space kg\) (mass of the pin), \(u_2 = 0\space m/s\) (initial velocity of the pin), \(v_1=+5.40\space m/s\) (final velocity of the ball), and \(v_2\) is the final velocity of the pin (which we need to find).
Substitute the values into the formula: \((7.30\times6.00)+(1.60\times0)=(7.30\times5.40)+(1.60\times v_2)\)

Step2: Simplify the equation

First, calculate the left - hand side: \(7.30\times6.00 = 43.8\). The right - hand side: \(7.30\times5.40=39.42\). So the equation becomes \(43.8=39.42 + 1.60v_2\)

Step3: Solve for \(v_2\)

Rearrange the equation to isolate \(v_2\): \(1.60v_2=43.8 - 39.42\). Then \(1.60v_2 = 4.38\). So \(v_2=\frac{4.38}{1.60}=2.7375\approx2.7\space m/s\)

Answer:

+2.7 m/s