A 65.5-kg person throws a 0.0495-kg snowball forward with a ground speed of 34.0 m/s. A second person, with a mass of 59.5 kg, catches the snowball. Both people are on skates. The first person is initially moving forward with a speed of 2.45 m/s, and the second person is initially at rest. What are the velocities of the two people after the snowball is exchanged? Disregard the friction between the skates and the ice. (Take the direction the snowball is thrown to be the positive direction. Indicate the direction with the sign of your answer.) thrower catcher m/s (Give your answer to at least three decimal places.) m/s
The Correct Answer and Explanation is :
Solution
We solve this problem using the principle of conservation of momentum, which states that the total momentum of the system before the exchange must equal the total momentum after the exchange because there are no external forces.
Step 1: Initial Momentum of the System
- Mass of the thrower (Person 1): m1=65.5 kgm_1 = 65.5 \, \text{kg}
- Initial velocity of the thrower: v1i=2.45 m/sv_{1i} = 2.45 \, \text{m/s}
- Mass of the snowball: ms=0.0495 kgm_s = 0.0495 \, \text{kg}
- Velocity of the snowball (relative to the ground): vs=34.0 m/sv_s = 34.0 \, \text{m/s}
- Mass of the catcher (Person 2): m2=59.5 kgm_2 = 59.5 \, \text{kg}
- Initial velocity of the catcher: v2i=0 m/sv_{2i} = 0 \, \text{m/s}
The initial momentum of the system: pinitial=m1v1i+msv1i+m2v2ip_{\text{initial}} = m_1 v_{1i} + m_s v_{1i} + m_2 v_{2i}
Substituting the values: pinitial=(65.5)(2.45)+(0.0495)(2.45)+(59.5)(0)p_{\text{initial}} = (65.5)(2.45) + (0.0495)(2.45) + (59.5)(0) pinitial=160.475+0.121275+0=160.596275 kg\cdotpm/sp_{\text{initial}} = 160.475 + 0.121275 + 0 = 160.596275 \, \text{kg·m/s}
Step 2: Final Momentum of the System
Let the final velocities be:
- Thrower’s velocity: v1fv_{1f}
- Catcher’s velocity: v2fv_{2f}
The final momentum of the system: pfinal=m1v1f+msvs+m2v2fp_{\text{final}} = m_1 v_{1f} + m_s v_s + m_2 v_{2f}
Using conservation of momentum: pinitial=pfinalp_{\text{initial}} = p_{\text{final}} 160.596275=(65.5)v1f+(0.0495)(34.0)+(59.5)v2f160.596275 = (65.5)v_{1f} + (0.0495)(34.0) + (59.5)v_{2f}
Simplify: 160.596275=(65.5)v1f+1.683+(59.5)v2f160.596275 = (65.5)v_{1f} + 1.683 + (59.5)v_{2f} 160.596275−1.683=65.5v1f+59.5v2f160.596275 – 1.683 = 65.5v_{1f} + 59.5v_{2f} 158.913275=65.5v1f+59.5v2f(1)158.913275 = 65.5v_{1f} + 59.5v_{2f} \tag{1}
Step 3: Snowball Exchange
- Thrower’s loss of snowball momentum: The thrower loses momentum equal to msvsm_s v_s.
- Catcher’s gain of snowball momentum: The catcher gains momentum equal to msvsm_s v_s.
Thus, after the throw: v1f=v1i−msvsm1v_{1f} = v_{1i} – \frac{m_s v_s}{m_1}
Substitute values: v1f=2.45−(0.0495)(34.0)65.5v_{1f} = 2.45 – \frac{(0.0495)(34.0)}{65.5} v1f=2.45−0.0257≈2.424 m/sv_{1f} = 2.45 – 0.0257 \approx 2.424 \, \text{m/s}
Step 4: Solve for Catcher’s Final Velocity
Substitute v1f=2.424v_{1f} = 2.424 into Equation (1): 158.913275=(65.5)(2.424)+(59.5)v2f158.913275 = (65.5)(2.424) + (59.5)v_{2f} 158.913275=158.772+(59.5)v2f158.913275 = 158.772 + (59.5)v_{2f} 158.913275−158.772=(59.5)v2f158.913275 – 158.772 = (59.5)v_{2f} 0.141275=(59.5)v2f0.141275 = (59.5)v_{2f} v2f=0.14127559.5≈0.0024 m/sv_{2f} = \frac{0.141275}{59.5} \approx 0.0024 \, \text{m/s}
Final Answers
- Thrower’s velocity after the throw: v1f≈2.424 m/sv_{1f} \approx 2.424 \, \text{m/s}
- Catcher’s velocity after catching the snowball: v2f≈0.0024 m/sv_{2f} \approx 0.0024 \, \text{m/s}
Explanation
The conservation of momentum ensures that the total momentum of the system remains constant. When the thrower throws the snowball, they lose momentum, while the catcher gains it after catching the snowball. The changes in velocity are relatively small due to the significantly larger masses of the individuals compared to the snowball.