Question

a 50 - kg student sitting in a rolling chair at rest pushes against a wall, which applies a 10 n - s horizontal impulse to the student. later, a 40 - kg student is at rest in the same rolling chair and catches a 10 - kg ball while applying a 10 n - s horizontal impulse to the ball. which of the following statements best describes the students final speeds and provides valid reasoning supporting the claim?
a both students have the same final speed because the 50 kg student - chair system and the 10 kg ball - 40 kg student - chair system have the same mass and experience impulses of the same magnitude.
b both students have the same final speed because each ball - chair - student system experiences the same impulse regardless of the mass of the system.
c the 40 - kg student has a greater final speed because the 40 - kg student - chair system has less mass than the 50 - kg student - chair system.
d the 40 - kg student has a greater final speed because the ball transfers all of its momentum to the student when it is caught.

Explanation:

Step1: Use the impulse - momentum theorem

The impulse - momentum theorem is \(J=\Delta p = m\Delta v\). For the first student (\(m_1 = 50\space kg\)), \(J_1=10\space N\cdot s\). Since the initial velocity \(v_{i1}=0\), we have \(J_1=m_1v_{f1}\), so \(v_{f1}=\frac{J_1}{m_1}\).

Step2: Analyze the second situation

For the second case, when the \(40 - kg\) student catches the \(10 - kg\) ball. According to the impulse - momentum theorem, the impulse on the student - chair - ball system (the impulse on the system is due to the internal force between the student and the ball, and the external impulse on the system is considered. Here, using the fact that for the student - chair - ball system, the impulse \(J_2 = 10\space N\cdot s\) (same magnitude as \(J_1\)). The mass of the second system \(m_2=40 + 10=50\space kg\) (but wait, no! Wait, for the second case, when calculating the velocity of the student - chair (after catching the ball), we should consider the impulse on the student - chair. The impulse on the student - chair is \(J = 10\space N\cdot s\) (from the ball - chair interaction, using Newton's third law, the impulse on the student - chair is equal in magnitude to the impulse on the ball). The mass of the student - chair system \(m_{sc}=40\space kg\). Using \(J = m_{sc}v_{f2}\), so \(v_{f2}=\frac{J}{m_{sc}}\)

For the first student (student - chair system, \(m_1 = 50\space kg\)), \(v_{f1}=\frac{J}{m_1}=\frac{10}{50}=0.2\space m/s\). For the second student (student - chair system, \(m_2 = 40\space kg\)), \(v_{f2}=\frac{J}{m_2}=\frac{10}{40}=0.25\space m/s\)

Answer:

C. The 40 - kg student has a greater final speed because the 40 - kg student - chair system has less mass than the 50 - kg student - chair system.