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 $60^{\circ}$ with the vertical, the tension in the string is $\left(g=10\, m / s ^{2}\right)$

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

Answer 1

Answer: (B)

From law of conservation of momentum, we known,

$m_{1} u_{1}+m_{2} u_{2}=m_{1} v_{1}+m_{2} v_{2}$
$u_{1}=0, u_{2}=150\, m / s , m_{1}=2.9\, kg$ and $m_{2}=0.1\, kg$
So, $2.9 \times 150=(2.9+0.1) v$
$\Rightarrow \frac{2.9 \times 150}{3}=v$
$\Rightarrow v=145\, m / s$
Also, $T \sin \theta=\frac{m v^{2}}{r}$
Putting the values and solving, we get $T=135\, N$.