Question

a mass of 100 g strikes the wall with speed 5m/s at an angle as shown in Figure and it rebounds with the same speed. if the contact tie is 2×10^-3 , what is the force applied​

Answer 1

Answer: ≈ 43.3 N

Explanation:

We use the Impulse-Momentum theorem to find the solution.

Clearly, the y-component of velocity remains unchanged before and after the collision ⇒ Δv(y) = 0

The change in velocity takes place only along the x-direction. Initially, the body moves towards the positive x-axis with velocity

v(x1)=v·sin(60°) = 5 x sin(60°)

After the brief collision, the magnitude of velocity remains unchanged but the direction is reversed completely, i.e.

v(x2) = - 5 x sin(60°)

Thus, change in velocity along x is

Δv(x) = 2 x 5 x sin(60°) = 10 x sin(60°)

From Impulse-Momentum theorem.

$F = \frac{dp}{dt} = m\frac{dv}{dt}$

(Apologies for the clumsy mathematical notation, I can't enter delta symbols into fractions!)

Thus,

F = Mass x Change in velocity/ Time = 10^(-1) x 10 x sin(60°)/2 x 10^(-3) N

F = 43.3 N

Answer 2

Explanation:

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