Question

A 2140 kgrailroad flatcar, which can move with negligible friction, is motionless next to a platform. A 242 kgsumo wrestler runs at 5.3 m/salong the platform (parallel to the track) and then jumps onto the flatcar. What is the speed of the flatcar if he then (a) stands on it, (b) runs at 5.3 m/s relative to it in his original direction, and (c) turns and runs at 5.3 m/s relative to the flatcar opposite his original direction?

Short answer

a) Speed of the flatcar if he stands on it, $vis0.54m/s$.

b) Speed of the flatcar if he runs at 5.3 m/s relative to the flatcar in the same direction, V is 0 m/s .

c) Speed of the flatcar if he turns and runs at 5.3 m/s relative to the flatcar in the opposite direction is V is 1.1 m/s .

Step-by-step solution

Understanding the given information

i) Mass of the flatcar, M is 2140 kg .

ii) Mass of sumo wrestler, m is 242 kg .

iii) The initial speed of the sumo wrestler,$v_{i1}is5.3m/s$ .

Concept and formula used in the given question

You use the concept of conservation of momentum. Using the given values, you can find the speed of the flatcar. The formula used is given below.

$m_1{v}_{i1}+m_2{v}_{i2}=m_1{v}_{f1}+m_1{v}_{f2}$

(a) Calculation for the speed of the flatcar if he then stands on it

Initially, the flatcar is at rest. We use the equation of conservation of momentum we get,

_{$m{v}_{i1}+M{v}_{i2}=m{v}_{f1}+M{v}_{f2}$ (1)}

Here$v_{i2}$ is the initial velocity of mass M, $v_{f2}$and$v_{f1}$ is the final velocity of mass M and m.

Since finally, both move together with a velocity v, let's substitute the values in the above expression, and we get,

$$\begin{gathered} 242\left(5.3\right)+M\left(0\right)=242v+2140v \\ 1282.6=2382v \\ v=0.54m/s \end{gathered}$$

Thus, the speed of the flatcar if he stands on it, $vis0.54m/s$.

(b) Calculation for the speed of the flatcar if he then runs at 5.3 m/s relative to it in his original direction

Here, the relative speed of the sumo is V + 5.3 m/s_{, then from}equation 1, we can calculate the speed of flatcar V as,

$$\begin{gathered} 242\left(5.3\right)+M\left(0\right)=242\left(V+5.3\right)+2140V \\ 1282.6=2382V+1282.6 \\ 2382V=0 \\ V=0m/s \end{gathered}$$

Thus, the speed of the flatcar if he runs at 5.3 m/s relative to the flatcar in the same direction, V is 0 m/s .

(c) Calculation for the speed of the flatcar if he then turns and runs at 5.3 m/s relative to the flatcar opposite his original direction

Here, the direction of sumo is opposite to the original direction; therefore, the new speed will be$V-v_{j1}$

From equation 1, we can calculate the speed of a flat car as,

$$\begin{gathered} 242\left(5.3\right)+M\left(0\right)=242\left(V-5.3\right)+2140V \\ 1282.6=242V-1282.6+2140V \\ 2565.2=2382V \\ V=1.1m/s \end{gathered}$$

Thus, the speed of the flatcar if he turns and runs at 5.3 m/s relative to the flatcar in the opposite direction is V is 1.1 m/s .