Question

A mass of 100g strikes the wall with speed 5m/s at an angle of 60 and rebounds with the same speed. If the contact time is 2*10^-3 sec., what is the force applied by the wall?

Answer 1

Given:

Mass = 100 g

Speed = 5 m / s

Angle = 60 degree

Contact time = 2* 10^-3 s

To find:

The force applied by the wall.

Solution:

To find the force,

Initial momentum, P1 = m v sin Θ i + m v cos Θ j

Final momentum, P2 = - m v sin Θ i + m v cos Θ j

Where,

P1, P2 - Initial and final momentum

m - mass

v - speed

Therefore,

Force = Change in momentum / Change in time

ΔP / ΔT

-2 m v sin Θ / 2 * 10^-3

Substituting the values,

We get,

-2 ( 0.1 ) ( 5 ) sin 60 / 2 * 10^-3

Force = -250 √3 N

The negative sign indicated the opposite direction of the force.

Hence, the force applied by the wall is -250 √3 N

Answer 2

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