91Ó°ÊÓ

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,$v_{i1}is5.3m/s$ .

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.

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

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, $vis0.54m/s$.

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 .

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 .