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Question: A ball reaches a racket at \(60\;{\text{m}}/{\text{s}}\) along +x direction, and leaves the racket i...

A ball reaches a racket at 60  m/s60\;{\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  gm50\;{\text{gm}} and the contact time is 0.02  s0.02\;{\text{s}}. With the force exerted by the racket on the ball, which of the following can be done? (Take g=10  m/s2g = 10\;{\text{m}}/{{\text{s}}^2})
A. We can lift a load 300  kg300\;{\text{kg}} off the ground
B. We can lift a load 30  kg30\;{\text{kg}} child off the ground
C. We can lift a load 90  kg90\;{\text{kg}}stone off the ground
D. We can lift a load 300  tonnes300\;{\text{tonnes}} truck off the ground

Explanation

Solution

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  m/su = 60\;{\text{m}}/{\text{s}}
The velocity of the racket in opposite direction is v=−60  m/sv = - 60\;{\text{m}}/{\text{s}}
The mass of the ball is m=50  gm=50  gm×10−3  kg1  gm=0.050  kgm = 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  st = 0.02\;{\text{s}}
The value of the gravitational acceleration is g=10  m/s2g = 10\;{\text{m}}/{{\text{s}}^2}
The formula to calculate the momentum of the ball is given as,
Pb=mu{P_b} = mu
The formula to calculate the momentum of the racket is given as,
Pr=mv{P_r} = mv
The formula to calculate the impulse of the ball is given as,
I=F⋅tI = F \cdot t
Apply the impulse-momentum theorem to find the force exerted by the ball on the racket.
Pb−Pr=I{P_b} - {P_r} = I
⇒mu−mv=F⋅t\Rightarrow mu - mv = F \cdot t
⇒m(u−v)=F⋅t\Rightarrow m\left( {u - v} \right) = F \cdot t
⇒F=m(u−v)t......(1)\Rightarrow F = \dfrac{{m\left( {u - v} \right)}}{t}......\left( 1 \right)
Substitute 0.050  kg0.050\;{\text{kg}}for m, 60  m/s60\;{\text{m}}/{\text{s}}for u, −60  m/s - 60\;{\text{m}}/{\text{s}} for v and 0.02  s0.02\;{\text{s}}in the expression (1) to find the force exerted by the ball on racket.
F=(0.050  kg)(60  m/s−(−60  m/s))0.02  sF = \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  NF = 300\;{\text{N}}
The formula to calculate the mass that can exert the same effect as exerted by the ball is given as,
M=Fg......(2)M = \dfrac{F}{g}......\left( 2 \right)
Substitute 300  N300\;{\text{N}} for F and 10  m/s210\;{\text{m}}/{{\text{s}}^2}for gg to find the mass that can exert same effect as exerted by the ball.
M=300  N10  m/s2M = \dfrac{{300\;{\text{N}}}}{{10\;{\text{m}}/{{\text{s}}^2}}}
∴M=30  kg\therefore M = 30\;{\text{kg}}

Thus, the force exerted by the ball can lift the 30  kg30\;{\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.