Question

3.2.2 quiz: momentum
question 1 of 10
a 58.72 kg person holding a steel ball stands motionless on a frozen lake. the person then throws the ball, which propels the person at 1.05 m/s to the right and the ball 3.75 m/s to the left. if the initial momentum of the system is zero, what is the mass of the steel ball?
a. 3.57 kg
b. 16.44 kg
c. 3.73 kg
d. 15.66 kg

Explanation:

Step1: Apply law of conservation of momentum

Since initial momentum is 0, final momentum of person and ball must sum to 0. Let mass of person be $m_p = 58.72$ kg, velocity of person be $v_p=1.05$ m/s, mass of ball be $m_b$ and velocity of ball be $v_b = - 3.75$ m/s (negative as it moves in opposite direction). So, $m_pv_p+m_bv_b = 0$.

Step2: Solve for mass of ball

Rearrange the equation $m_pv_p+m_bv_b = 0$ to $m_b=\frac{-m_pv_p}{v_b}$. Substitute $m_p = 58.72$ kg, $v_p = 1.05$ m/s and $v_b=-3.75$ m/s into the formula: $m_b=\frac{-58.72\times1.05}{-3.75}=\frac{- 61.656}{-3.75}=16.44$ kg.

Answer:

B. 16.44 kg