# A ball of mass 'm' moving with a speed 'u' along a direction making an angle α with the vertical strikes a horizontal steel plate. The average force exerted by the plate on the ball is?

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by varshamonala

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by varshamonala

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It is not given whether the horizontal steel plate is a very heavy one and it does not move like the floor , or it also moves after being hit by the ball. It is not given if the collision is an elastic one or not. It is not clear, if it is an elastic collision , how long the ball is in contact with the plate.

The momentum of the ball before collision = m u (Sin α**i** - Cos α **j **)

__Inelastic collision__ :

it is not clear, in how much time the ball came to rest after hitting the plate. It is not clear if the plate moved at all.

Lose of Momentum = - m*u

force = m * u (Cos α**j** - Sin α **i**) / (time duration for the ball to come to rest) + m g **j**.

direction of the force is resultant of the two components of the forces.

after the ball comes to rest, the force is just mg**j**.

__Elastic collision__:

we assume the ball bounces back making angle of α with the vertical.

momentum after collision = m u (Sin α** i** + Cos α** j )**

change in momentum = 2 m u Cos α** j** .

Force on the ball by the plate = 2 m u Cos α**j** / time duration of contact + m g **j**.

The momentum of the ball before collision = m u (Sin α

it is not clear, in how much time the ball came to rest after hitting the plate. It is not clear if the plate moved at all.

Lose of Momentum = - m*u

force = m * u (Cos α

direction of the force is resultant of the two components of the forces.

after the ball comes to rest, the force is just mg

we assume the ball bounces back making angle of α with the vertical.

momentum after collision = m u (Sin α

change in momentum = 2 m u Cos α

Force on the ball by the plate = 2 m u Cos α