Two masses 8 kg and 12 kg are connected at the two ends of a light inextensible string that goes over a frictionless pulley. Find the acceleration of the masses and the tension in the string when the masses are released.

Explanation

Let `m_(1) = 8 kg m_(2) = 12 kg`
Let tension in string be T

Let common acceleration of system is a
Acceleration of `m_(1)` is in upward direction acceleration of `m_(2)` is in downward direction equation of motion of `m_(1)`
By taking addition of equation (1) and (2) we get common acceleration of system
`(m_(2) - m_(1)) g= (m_(1) + m_(2))a`
`=((12-8)/( 8+12)) xx 10`
`T = m_(1) g + m_(1) ((m_(2)-m_(1))/( m_(1)+ m_(2))) g`
`T= ((2m_(1)m_(2))/( m_(1)+m_(2)))g`
Substituting value
`T= (2xx 8 xx 12)/( 8+ 12) xx 10`
`T= (2xx 960)/( 20)`
`:. T = 96 N`
Let mass of nucleus is m and initial velocity v= 0
initial momentum `vec(p )_(1) =- mV =0`
After disintergration it is divided into two part let mass be `m_(1)` and `m_(2)` and velocity be `vec( v ) _(1)` and `vec(2)` respectively
By law of conservation of linear momentum
`vec(p )_(1) = vec( p)_(f)`
`:. ,m_(2) v_(2) = - m_(1) m_(1)`
Negative sigh shows that velocity of both fragment are in mutually opposite direction