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Classical Mechanics Level 4

A gun which fires small balls of mass 20 g is firing 20 balls per second on the smooth horizontal table surface ABCD . If the collision is perfectly elastic and balls are striking at the centre of table with a speed of 5m s − 1 s^{-1} at an angle of 60° with the vertical just before collision ,then force exerted by one of the legs on ground is ?

(Assume total weight of the table is 0.2kg)

0.75N 0.25N 0.5N 1N

Prashant Kr by Prashant Kr , 22, India

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1 solution

Prashant Kr
Apr 22, 2015

Force on the table due to collision of balls :

F d y n a m i c F_{dynamic} = d p d t \frac{dp}{dt} =2 x 20 x 20 x 1 0 − 3 10^{-3} x 5 x 0.5=2N

net force on one leg = 1 4 \frac{1}{4} (2+0.2 x 10) =1 N

1 Helpful 0 Interesting 0 Brilliant 0 Confused

but mg+1=4N

n=force by 1 ieg

m=0.2kg

so N=0.75N

Akash singh - 5 years, 8 months ago

Please let us know that the table is on the earth. Question is a bit smoggy.

Kaushik Chandra - 3 years, 11 months ago

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