Question

. A ball of mass 100 kg collides with a wall while moving
with a horizontal velocity of 10 m/s and return with
velocity 8m/s if the collision time is 0.1 s then find the
average impulsive force exerted by the wall on the ball.​

Answer 1

Answer:

• The Magnitude of impulsive force exerted by the wall on the ball = 18000 N

Explanation:

Given:

• Mass of ball, m = 100 kg
• Initial velocity of ball before collision, u = 10 m/s
• Final velocity of ball after collision, v = -8 m/s  

[ negative final velocity because ball returns back ]

• time of collision, t = 0.1 s

To find:

• The Average impulsive force exerted by the wall on the ball, F =?

Knowledge required:

Simply, the force acting on an object for a small duration of times is called impulsive force.

So,

• Formula to calculate the impulsive force (F)

        F = (Final momentum - Initial momentum) / time

• Formula to calculate momentum

        Momentum = mass × velocity  

Solution:

Using the formula

→ F = ( m v - m u ) / t

→ F = [ (100) (-8) - (100) (10) ] / (0.1)

→ F = ( - 800 - 1000 ) / 0.1

→ F = -1800 / 0.1

→ F = -18000 N

Therefore,

• The magnitude of the impulsive force exerted by the wall on the ball is 18000 Newtons.

Answer 2

Answer:

✯ $\large{\red{\bold{\underline{GIVEN :-}}}}$

• A ball of mass 100 kg collides with a wall while moving with a horizontal velocity of 10 m/s and return with velocity of 8 m/s.
• The collision time is 0.1 s.

✯ $\large{\red{\bold{\underline{TO\: FIND :-}}}}$

• What is the average impulsive force exerted by the wall on the ball.

✯ $\large{\red{\bold{\underline{FORMULA\: USED :-}}}}$

$\small{\bf{\underline{\underline{❖\: Force\: =\: \dfrac{Final\: Momentum\: -\: Initial\: Momentum}{Time}}}}}$

✯ $\large{\red{\bold{\underline{SOLUTION :-}}}}$

⋆ Given :

• Mass (m) = 100 kg
• Initial Velocity (u) = 10 m/s
• Final Velocity (v) = - 8 m/s [For return back the final velocity will be negetive]
• Time (t) = 0.1 seconds

➣ According to the question by using the formula we get,

➛ F = $\dfrac{(100)(- 8) - (100)(10)}{0.1}$

➛ F = $\dfrac{- 800 - 1000}{0.1}$

➛ F = $\dfrac{- 1800}{0.1}$

➥ $\bf{\green{\underbrace{\blue{F\: =\: -\: 18000\: N}}}}$

$\therefore$ The average impulsive force exerted by the wall on the ball is $\boxed{\bold{\small{ 18000\: N}}}$

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