Question

A ball reaches a racket at $60\;{\text{m}}/{\text{s}}$ along +x direction, and leaves the racket in the opposite direction with the same speed. Assume that the mass of the ball is $50\;{\text{gm}}$ and the contact time is $0.02\;{\text{s}}$. With the force exerted by the racket on the ball, which of the following can be done? (Take $g = 10\;{\text{m}}/{{\text{s}}^2}$)
A. We can lift a load $300\;{\text{kg}}$ off the ground
B. We can lift a load $30\;{\text{kg}}$ child off the ground
C. We can lift a load $90\;{\text{kg}}$stone off the ground
D. We can lift a load $300\;{\text{tonnes}}$ truck off the ground

Answer 1

Hint:The above problem is based on the impulse-momentum theorem. The momentum describes the inertia of the moving object and momentum describes the effect of the force on the object for a given time interval. The impulse-momentum theorem states that the change in the momentum of the particle is equal to the impulse of the force.

Complete step by step answer:
Given: The velocity of the ball is $u = 60\;{\text{m}}/{\text{s}}$
The velocity of the racket in opposite direction is $v = - 60\;{\text{m}}/{\text{s}}$
The mass of the ball is $m = 50\;{\text{gm}} = 50\;{\text{gm}} \times \dfrac{{{{10}^{ - 3}}\;{\text{kg}}}}{{1\;{\text{gm}}}} = 0.050\;{\text{kg}}$
The time for the contact of the ball with racket is $t = 0.02\;{\text{s}}$
The value of the gravitational acceleration is $g = 10\;{\text{m}}/{{\text{s}}^2}$
The formula to calculate the momentum of the ball is given as,
${P_b} = mu$
The formula to calculate the momentum of the racket is given as,
${P_r} = mv$
The formula to calculate the impulse of the ball is given as,
$I = F \cdot t$
Apply the impulse-momentum theorem to find the force exerted by the ball on the racket.
${P_b} - {P_r} = I$
$\Rightarrow mu - mv = F \cdot t$
$\Rightarrow m\left( {u - v} \right) = F \cdot t$
$\Rightarrow F = \dfrac{{m\left( {u - v} \right)}}{t}......\left( 1 \right)$
Substitute $0.050\;{\text{kg}}$for m, $60\;{\text{m}}/{\text{s}}$for u, $ - 60\;{\text{m}}/{\text{s}}$ for v and $0.02\;{\text{s}}$in the expression (1) to find the force exerted by the ball on racket.
\[F = \dfrac{{\left( {0.050\;{\text{kg}}} \right)\left( {60\;{\text{m}}/{\text{s}} - \left( { - 60\;{\text{m}}/{\text{s}}} \right)} \right)}}{{0.02\;{\text{s}}}}\]
\[F = 300\;{\text{N}}\]
The formula to calculate the mass that can exert the same effect as exerted by the ball is given as,
$M = \dfrac{F}{g}......\left( 2 \right)$
Substitute \[300\;{\text{N}}\] for F and $10\;{\text{m}}/{{\text{s}}^2}$for $g$ to find the mass that can exert same effect as exerted by the ball.
\[M = \dfrac{{300\;{\text{N}}}}{{10\;{\text{m}}/{{\text{s}}^2}}}\]
\[\therefore M = 30\;{\text{kg}}\]

Thus, the force exerted by the ball can lift the $30\;{\text{kg}}$ child off the ground and option (B) is the correct answer.

Note:Convert mass into SI unit before substituting in the formula. Remember that the speed of the racket will be the same as the velocity of the ball in the opposite direction. Substitute the value of gravitational acceleration as given in the problem.