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A 3.00 kg steel ball strikes a massive wall at 10.0m/s at
anangle of 60.0 degree with the plane of the wall. It bouncesoff
the wall with the same speed and angle. If the ball is incontact
with the wall for 0.200s, what is the average force exertedby the
wall on the ball?

Respuesta :

Answer:259.80 N

Explanation:

Given

mass of ball=3 kg

ball velocity =10 m/s

angle made by ball with plane of wall [tex]\theta [/tex]=60

Momentum change in Y direction remains same and there is change only in x direction

therefore

initial momentum[tex]=mvsin\theta [/tex]

=30sin60

Final momentum=-30sin60

Change in momentum is =30sin60+30sin60

=60sin60

and Impulse = change in momentum

Fdt=dP

where F=force applied

dP=change in momentum

[tex]F\times 0.2=60sin60[/tex]

[tex]F\times 0.2=51.96[/tex]

F=259.80 N

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