Question

. During a collision with a wall, the velocity of a 0.200-kg ball changes from 20.0 m/s toward the wall to 12.0 m/s away from the wall. If the time the ball was in contact with the wall was 60.0 ms, what was the magnitude of the average force applied to the ball?

Answer 1

The magnitude of the average force applied to the ball was about 107 N

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Further explanation

Newton's second law of motion states that the resultant force applied to an object is directly proportional to the mass and acceleration of the object.

$\large {\boxed {F = ma }$

F = Force ( Newton )

m = Object's Mass ( kg )

a = Acceleration ( m )

Let us now tackle the problem !

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Given:

initial velocity of the ball = u = -20.0 m/s

final velocity of the ball = v = 12.0 m/s

contact time = t = 60.0 ms = 0.06 s

mass of the ball = m = 0.200 kg

Asked:

average force applied to the ball = F = ?

Solution:

We will use this following formula to solve this problem:

$\Sigma F = ma$

$F = m ( v - u ) \div t$

$F = 0.200 ( 12 - (-20) ) \div 0.06$

$F = 0.200 ( 32 ) \div 0.06$

$F = 6.4 \div 0.06$

$F = 106 \frac{2}{3} \texttt{ N}$

$F \approx 107 \texttt{ N}$

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Learn more

• Impacts of Gravity : brainly.com/question/5330244
• Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
• The Acceleration Due To Gravity : brainly.com/question/4189441
• Newton's Law of Motion: brainly.com/question/10431582
• Example of Newton's Law: brainly.com/question/498822

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Answer details

Grade: High School

Subject: Physics

Chapter: Dynamics

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Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant

Answer 2

Answer:

106.7 N

Explanation:

We can solve the problem by using the impulse theorem, which states that the product between the average force applied and the duration of the collision is equal to the change in momentum of the object:

$F \Delta t = m (v-u)$

where

F is the average force

$\Delta t$ is the duration of the collision

m is the mass of the ball

v is the final velocity

u is the initial velocity

In this problem:

m = 0.200 kg

u = 20.0 m/s

v = -12.0 m/s

$\Delta t = 60.0 ms = 0.06 s$

Solving for F,

$F=\frac{m(v-u)}{\Delta t}=\frac{(0.200 kg) (-12.0 m/s-20.0 m/s)}{0.06 s}=-106.7 N$

And since we are interested in the magnitude only,

F = 106.7 N