First slide

Vertical circular motion

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^o with the vertical, the tension in the string is (g=10m/s²)

Moderate

A

140 N
B

135 N
C

125 N
D

190 N

Solution

Conserving Linear momentum,
$\large (2.9+0.1)v{}'=0.1\times150\Rightarrow{v}'=5m/s$

Conserving energy between A and B,
$\large \frac 12mv^2+mgl(1-cos\theta)=\frac 12m({v}')^2$

$\large \Rightarrow v^2=(5)^2-2\times10\times0.5\times(1-cos60^0)$

$v^{2} = 20$

$\large \therefore T=mgcos\theta+\frac {mv^2}{l}$

$\large =\left ( 3\times 10\times cos60^0+\frac {3\times 20}{0.5} \right )N\Rightarrow T=135N$

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