An 8.18-kg point mass and a 16.6-kg point mass are held in place 50.0 cm apart. A particle of mass m is released from a point between the two masses 20.0 cm from the 8.18-kg mass along the line connecting the two fixed masses. Find the magnitude and direction of the acceleration of the particle.
magnitude | m/s2 |
Force exerted by one body having mass m1 on another having mass m2 and distance between them is r ,is given by
( where G is gravitational constant )
In given question a body of mass m is kept between two different bodies , hence both the bodies having masses 8.18 kg and 16.6 kg will exert force and try to attract the body with mass m.
Let force exerted by 8.18 kg mass body is F1 and distance between them is given 20 cm . or 20x10-2 meter.
F1= 1364.01 x10-11 N ( towards itself)
Now let force exerted by 16.6 kg body is F2. Distance of m mass body is (50-20)= 30 cm or 30x10-2 meter
F2=1230.24x10-11 N ( towards itself)
Both force F1 and F2 are acting on same body and in opposite direction hence net fore F is given by
F= (F1) -(F2)
F=(1364.01 x10-11 ) - (1230.24x10-11)
F=133.77x10-11 N ( towards 8.18 kg body)
Let acceleration of body with mas m is .
F= m
= F/m
( towards 8.18 kg mass body) (ANS)
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