A mass of 2.9 kg is suspended from a string of length 50 cm and is at rest. Another body of mass 100 g, which is moving horizontally with a velocity of `150 m//s` strikes and sticks to it. Subsequently when the string makes an angle of `60^@` with the vertical, the tension in the string is (g = `10 m//s^(2)`)

Options

A. 140 N
B. 135 N
C. 125 N
D. 90 N

Answer

B

Explanation

From Law of conservation of momentum
`m_(1)+ m_(2)U_(2) = m_(1)V_(1)+ m_(2) V_(2)`
`2.9 xx 0 + 0.1 xx 150 = (2.9 + 0.1) V`
`v=5m//s`
For circular motion,
Centripetal force = Net force towards centre
`implies(m_(1)+m_(2)v_(1)^(2))/l=T-(m_(1)+m_(2))gcos60^@`
`implies T =3 xx 9.8 xx 1/2+(3v_(1)^(2))/0.5`
`= 14.7 + 6V_(1)^(2)`
According to Law of conservation of mechanical energy,
K.E at lowest point = (K.E + P.E)
`1/2(m_(1) + m_(2)) V^(2)= 1/2(m_(1) + m_(2)) V^_(1)(2)+ (m_(1) + m_(2))gh`
`(1/2)(3)(5)^(2) = 1/2(3)V_(1)^(2) +3xx9.8 xx0.5(1-1/2)`
`[because h= l(1- cos theta)= 0.5(1-cos 60^@)]`
`V_(1)^(2) = 20.1 (m//sec)`
On substituting `V_(2)` in equation (1),
`T= 14.7 + 6 xx 20.1 = 135.3 N`.