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Question: The \(100\,g\) of an iron ball having velocity \[10{\text{ }}m{s^{ - 1}}\] collides with a wall at a...

The 100 g100\,g of an iron ball having velocity 10 ms−110{\text{ }}m{s^{ - 1}} collides with a wall at an angle 30∘30^\circ and rebounds with the same angle. if the period of contact between the ball and wall is 0.1 second, then the force experienced by the ball is
A. 10 N
B. 100 N
C. 1 N
D. 0.1 N

Explanation

Solution

in order to solve the question, we will formula of rate of change in momentum and to apply that we will make the diagram so as to see the pictorial representation of resolving the components of momentum after the resolving we will apply the momentum in formula along with the time and get the answer

Formula used:
Force =rate of change in momentum
F=M×VTF = \dfrac{{M \times V}}{T}
Here, FF refers to force, MM refers to mass, VV refers to velocity and TT refers to time period.

Complete step by step answer:
In the question we are an iron ball which collides with a wall and rebounds with the same angle and we have to find the force experienced by the ball.
Mass of iron ball = 100 g
Velocity of iron ball = 10 ms−110{\text{ }}m{s^{ - 1}} collides with a wall
Ball collides the wall at an angle = 30∘30^\circ
period of contact between the ball and wall = 0.1 s

This diagram shows how the ball collides with the wall and then rebound at an angle of 30∘30^\circ . Now we resolve the vertical and horizontal component; the horizontal component will add up mvsin⁡θ+mvsin⁡θmv\sin \theta + mv\sin \theta while the component mvcos⁡θmv\cos \theta will cancel out.

This diagram shows the resolved component, now the change in momentum will be 2mvsin⁡θ2mv\sin \theta which is the product of mass and velocity. Applying the formula
F=M×VTF = \dfrac{{M \times V}}{T}
M×V=2mvsin⁡θM \times V = 2mv\sin \theta
θ=30∘\theta = 30^\circ and T = 0.1 s as given in the question

Now substituting the values in the equation
F=M×VTF = \dfrac{{M \times V}}{T}
⇒F=2mvsin⁡30∘0.1\Rightarrow F = \dfrac{{2mv\sin 30^\circ }}{{0.1}}
⇒F=2×0.1×10×sin⁡30∘0.1\Rightarrow F = \dfrac{{2 \times 0.1 \times 10 \times \sin 30^\circ }}{{0.1}}
We have converted mass from g to kg
⇒F=2×0.1×10×0.50.1N\Rightarrow F = \dfrac{{2 \times 0.1 \times 10 \times 0.5}}{{0.1}}N
⇒F=10 N\Rightarrow F = 10{\text{ }}N
Therefore, the force experienced by the ball is 10 N10{\text{ }}N.

Hence the correct option is A.

Note: Many of the students will make the mistake by not taking the physical quantities in their SI unit especially in this question we are given the mass of iron ball 100g and we have to convert into kg so that the force which comes out would remain in newton along with this we also have use diagram as to make question simple.