Question

A cricket ball of mass $150{\rm{ g}}$ moving with a velocity of $12{\rm{ }}{{\rm{m}} {\left/
 {\vphantom {{\rm{m}} {\rm{s}}}} \right.
} {\rm{s}}}$ strikes against the bat. It rebounces with a velocity of $20{\rm{ }}{{\rm{m}} {\left/
 {\vphantom {{\rm{m}} {\rm{s}}}} \right.
} {\rm{s}}}$. The ball remains in touch with the bat for $0.01{\rm{ s}}$. The average force applied by the bat on the ball is
A. $840{\rm{ N}}$
B. $480{\rm{ N}}$
C. $804{\rm{ N}}$
D. $408{\rm{ N}}$

Answer 1

Hint: In the solution we use the relation of the momentum of an object. The momentum of the object varies according to the mass and speed of the object.

Complete step by step answer:
Initial velocity of ball $u = 12{\rm{ }}{{\rm{m}} {\left/
 {\vphantom {{\rm{m}} {\rm{s}}}} \right.
} {\rm{s}}}$
Final velocity of ball $v = 20{\rm{ }}{{\rm{m}} {\left/
 {\vphantom {{\rm{m}} {\rm{s}}}} \right.
} {\rm{s}}}$
Time the ball is in touch with the bat $t = 0.01{\rm{ s}}$
Mass of the ball $m = 150{\rm{ g }}$
Converting the mass of ball into SI units gives $m = 150{\rm{ g}} \times \dfrac{{1{\rm{ kg}}}}{{1000{\rm{ g}}}} = 0.15{\rm{ kg}}$
Newton’s second law says that force is given by the rate of change of momentum
Change in momentum is given by
$\Delta p = m \times \left( {v - u} \right)$
Here the ball is hit by the bat and then travels in opposite direction, so the ball changes its direction and therefore, the initial velocity is taken as negative, that is $u = - 12{{{\rm{ m}}} {\left/
 {\vphantom {{{\rm{ m}}} {\rm{s}}}} \right.
} {\rm{s}}}$.
Substituting the values gives
$
\Delta p = 0.15{\rm{ kg}} \times \left( {20\;{{{\rm{ m}}} {\left/
 {\vphantom {{{\rm{ m}}} {\rm{s}}}} \right.
} {\rm{s}}} - \left( { - 12\;{{{\rm{ m}}} {\left/
{\vphantom {{{\rm{ m}}} {\rm{s}}}} \right.
} {\rm{s}}}} \right)} \right)\\
\Rightarrow\Delta p = 0.15\;{\rm{kg}} \times \left( {20\;{{{\rm{ m}}} {\left/
{\vphantom {{{\rm{ m}}} {\rm{s}}}} \right.
} {\rm{s}}} + 12\;{{{\rm{ m}}} {\left/
{\vphantom {{{\rm{ m}}} {\rm{s}}}} \right.
} {\rm{s}}}} \right)\\
\Rightarrow\Delta p = 4.8{{{\rm{ kg}} \cdot {\rm{m}}} {\left/
{\vphantom {{{\rm{ kg}} \cdot {\rm{m}}} {\rm{s}}}} \right.
} {\rm{s}}}
$
The relation between force, change in momentum and time is given by
$F = \dfrac{{\Delta p}}{t}$
Substituting the values in the above equation gives
$
F = \dfrac{{4.8{\rm{ }}{{{\rm{kg}} \cdot {\rm{m}}} {\left/
{\vphantom {{{\rm{kg}} \cdot {\rm{m}}} {\rm{s}}}} \right.
} {\rm{s}}}}}{{0.01{\rm{ s}}}}\\
\therefore F = 480{\rm{ N}}
$
Therefore, the correct answer is option B that is $480{\rm{ N}}$.

Note: Make sure to notice the change in direction of the ball as it affects the sign of the initial velocity and then the change in momentum.Moreover,momentum is a measurement of mass in motion: how much mass is in how much motion.