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)

Question

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) (A) 140 N (B) 135 N (C) 125 N (D) 90 N

Tardigrade Admin · Accepted Answer

From law of conservation of momentum, we known,

m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}

u_{1} = 0, u_{2} = 150 m / s, m_{1} = 2.9 k g

and

m_{2} = 0.1 k g

So,

2.9 \times 150 = (2.9 + 0.1) v

\Rightarrow \frac{2.9 \times 150}{3} = v

\Rightarrow v = 145 m / s

Also,

T sin θ = \frac{m v ^{2}}{r}

Putting the values and solving, we get

T = 135 N

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Q. A mass of 2.9kg is suspended from a string of length 50cm and is at rest. Another body of mass 100g, which is moving horizontally with a velocity of 150m/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=10m/s2)

140 N

135 N

125 N

90 N

From law of conservation of momentum, we known, m1​u1​+m2​u2​=m1​v1​+m2​v2​ u1​=0,u2​=150m/s,m1​=2.9kg and m2​=0.1kg So, 2.9×150=(2.9+0.1)v ⇒32.9×150​=v ⇒v=145m/s Also, Tsinθ=rmv2​ Putting the values and solving, we get T=135N.