Amass of 2.9 kg is suspended from a string of length 50 cm and is at rest. Another body of mass 10 gm which is moving horizontally with a velocity of 150 m/sec strikes it. After striking the two bodies combine together. The tension in the string, when it makes an angle of 60° with the vertical, is:

Options

A. 135.3 N
B. 165.5 N
C. 142.4 N
D. 90 N

Answer

A

Explanation

Applying the law of conservation of linear momentum,
`m u = (M + m) V " or " V = (m u)/((M + m)) implies (0.1 xx 150)/((0.1 xx 2.9)) implies 5 m//sec`

If `V_1` is the speed of the combined mass of the instant when the string makes an angle `60^@` with the vertical, then,
`T = (M + m) g cos 60^@ + ((M + m)V_1^2)/(l) = 3 xx 9.8 xx 1/2 + ((3V_1^2)/(0.5))`
`= 14.7 + 6V_1^2 " " ....(1)`
According to law of conservation of mechanical -energy we have
`(K.E.)_("lowest point") implies (K.E+P.E)_("any other point")`
or `1/2 (m + M) V^2 = 1/2 (m + M) V_1^2 + (m + M) gh`
or `1/2 (m + M)V^2 implies 1/2(m + M) V_1^2 + (m + M) gl (1 - cos theta)`
or `,1/2 xx 3 xx 25 = 3/2 V_1^2 + 3 xx 9.8 xx 0.5 (1 - 1/2) " " ...(2)`
Solving for `V_1^2`, we get `V_1^2 = 20.1 m//s^2`
Putting this value in egn.(1), we get, `T = 14.7 + 6 xx 20.1 implies 135.3 N`