Question

Two skaters $(A and B),$ each of mass $70 kg,$ are approaching each another on a frictionless surface, each with a speed of $1 {ms}^{- 1} .$ Skater A carries a ball of mass $10 kg .$ Both skaters can toss the ball at $5 {ms}^{- 1}$ relative to themselves such that when $A$ tosses the ball at $t = 0 s$ to $B$ then the ball leaves at $6 {ms}^{- 1}$ with respect to the ground. Further, they start $(t = 0 s)$ tossing the ball back and forth when they are $10 m$ apart $(see figure (1)) .$ Assume that the motion is one dimensional, all collisions are completely inelastic and that the time delay between receiving the ball and tossing it back is $1 s .$

$(a)$ State initial momenta of skaters (just before $t = 0 s$ ).

${\vec{P}}_{A} =; {\vec{P}}_{B}$

$(b)$ At $t = 0 s$ skater $A$ tosses the ball to skater $B .$ State momenta of both the skaters immediately after $B$ catches the ball.

${\vec{P}}_{A} =; {\vec{P}}_{B}$

$(c)$ Indicates the minimum number of tosses by each skater required to avoid collision.

Number of tosses by $A =$ Number of tosses by $B =$

(d) Indicate motion of each skater on the following $x - t plot$ if no tosses are made. [Note : For this and next part you must select the scale on the time axis approximately. You may use a pencil for sketching].

$(e)$ Indicate motion of each skater on the following $x - t plot$ from $t = 0 s$ till just after one round trip by the ball (from A to B and back to A).

Accepted Answer

$(a)$ Initially the ball is held by person $A$ . Hence the total mass will be $70 + 10 = 80 kg$ and approach velocity is $1 {ms}^{- 1}$ . Hence momentum of $A$ is $p_{A} = 80 kg {ms}^{- 1} \Rightarrow \vec{p_{A}} = 80 \hat{i} kg {ms}^{- 1}$
and for person $B$ momentum is $\vec{p_{B}} = - 70 \hat{i} kg {ms}^{- 1}$

$(b)$ Now, when $A$ throws the ball towards $B$ , due to linear momentum conservation

$80 (1) = 70 v + 10 (6) \Rightarrow v = \frac{2}{7} m s^{- 1}$ .

Linear momentum of person $A$ after the throw is given as ${\vec{P}}_{A} = 70 \frac{2}{7} = 20 \hat{i} kg {ms}^{- 1}$ .

When $B$ catches the ball, it sticks to him/her. Applying linear momentum conservation, we can write

$10 (6) - 70 (1) = 80 V \Rightarrow V = \frac{- 1}{8} m s^{- 1}$ .

Linear momentum of $B =$ is ${\vec{P}}_{B} = 80 (\frac{- 1}{8}) = - 10 \hat{i} kg {ms}^{- 1}$ .

$(c)$ Just after throwing the ball, speed of person $A$ is $v = \frac{2}{7} {ms}^{- 1}$ .

Relative speed between the persons is $\frac{2}{7} + 1 = \frac{9}{7} = 1.29 m s^{- 1}$

Relative velocity between ball and person $B$ is $6 + 1 = 7 m s^{- 1}$

Time taken by ball to reach $B$ is given as $t = \frac{10}{7} = 1.43 s$

Speed of person $B$ after catching the ball is $V = \frac{1}{8} = 0.125 m s^{- 1}$

After catching the ball, person will throw it back to person $A$ after $1$ second. i.e. at $t = 2.43 s$

Velocity of person $B$ after he/she throws it back to $A$ is determined by apply momentum conservation.

$- 10 = - 10 (\frac{1}{8} + 5) + 70 V' \Rightarrow V' = \frac{33}{72} = 0.45 m s^{- 1}$ .

$(d)$ When no person throws th ball, they will move with constant velocities and their relative velocities will be $2 m s^{- 1}$ .

They will meet at time $t = \frac{10}{2} = 5 s$

$(e)$ As per our previous calculations, the plot between position time graph of person $A$ and person $B$ is given as

Important Questions on Centre of Mass, Momentum and Collisions

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