Question

A cricket ball of mass 65 g moving with a velocity of 0.6 m/s stopped by a player in 0.6 seconds. What is the force applied by the player to stop the ball?​

Answer 1

Mass of the ball=70 g

Initial Velocity(u)=0.5 m/s

Final velocity(v)=0

Change in momentum=m(v-u)=-70×0.5×10⁻³=0.035 kgm/s

Time taken=0.5 s

∴Force applied to stop the ball=change in momemtum/time taken=0.07 N

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Answer 2

Given :-

Mass of the cricket ball = 65 g

Initial velocity = 0.06 m/s

Time taken = 0.6 sec

To Find :-

The force applied by the player to stop the ball.

Analysis :-

Here we are given with the mass, initial velocity, time and the final velocity.

Firstly, find the acceleration by substituting the given values using the first law of motion.

Then find the force accordingly by substituting the values we got such that force is equal to mass multiplied by acceleration.

Solution :-

We know that,

• a = Acceleration
• f = Force
• u = Initial velocity
• t = Time
• v = Final velocity
• m = Mass

Using the formula,

$\underline{\boxed{\sf First \ equation \ of \ motion=v=u+at}}$

Given that,

Mass (m) = 65 g = 0.065 kg

Initial velocity (u) = 0.06 m/s

Time taken (t) = 0.6 sec

Final velocity (v) = 0 m/s

Substituting their values,

⇒ 0 = 0.06 + (a) × 0.6

⇒ a = (v-u)/t

⇒ a = (0-0.06)/0.6

⇒ a = -0.06/0.6

⇒ a = -0.1 m/s²

Using the formula,

$\underline{\boxed{\sf Force=Mass \times Acceleration}}$

Given that,

Mass (m) = 0.065 kg

Acceleration (a) = -0.1 m/s²

Substituting their values,

⇒ f = 0.065 × -0.1

⇒ f = -0.0065

Therefore, the force applied by the player to stop the ball is -0.0065 N.