Answer to Question #103286 in Mechanics | Relativity for Sam

Question

Two rollerbladers face each other and stand at rest on a flat parking lot. Tony has a mass of 65kg, and Tim has a mass of 52kg. when the push off against one another, Tim acquires a speed of 0.63 m/s. What is Tony’s speed?

Answer 1

We can find Tony's speed from the law of conservation of momentum. Let's choose the direction in which Tony moves as a positive and apply the law of conservation of momentum:

$$m_{Tony}v_{Tony,i} + m_{Tim}v_{Tim,i} = m_{Tony}v_{Tony,f} - m_{Tim}v_{Tim,f},$$

here, $m_{Tony} = 65kg$ is the mass of Tony, $m_{Tim} = 52kg$ is the mass of Tim, $v_{Tony,i} = 0 \dfrac{m}{s}$ is the initial speed of Tony, $v_{Tim,i} = 0 \dfrac{m}{s}$ is the initial speed of Tim, $v_{Tony,f}$ is the final speed of Tony and $v_{Tim,f} = 0.63 \dfrac{m}{s}$ is the final speed of Tim.

From this equation we can find the final speed of Tony:

$$m_{Tony}v_{Tony,f} - m_{Tim}v_{Tim,f} = 0,$$ $$m_{Tony}v_{Tony,f} = m_{Tim}v_{Tim,f},$$ $$v_{Tony,f} = \dfrac{m_{Tim}v_{Tim,f}}{m_{Tony}} = \dfrac{52kg \cdot 0.63 \dfrac{m}{s}}{65kg} = 0.504 \dfrac{m}{s}.$$

Answer:

$v_{Tony,f} = 0.504 \dfrac{m}{s}.$