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Standard XII

Physics

Acceleration in 2D

Masses m1 a...

Question

Masses $m_{1}$ and $m_{2}$ are connected by a light string in figure. The wedges is smooth and the pulley is smooth and fixed $m_{1} = 10 k g$ and $m_{2} = 7.5 k g$ . When $m_{2}$ is just released, the distance it will travel in 2 seconds is ( $g = 10 m / s^{2}$ )

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7.5m

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2.8m

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Solution

The correct option is C 2.8m
We can write two equations for motion of these blocks,

$m_{2} g - T = m_{2} a$
$T - m_{1} g s i n 30 = m_{1} a$

Putting the values we get a= $\frac{25}{17.5} m / s^{2}$

Now distance travelled by $m_{2}$ in two seconds = $\frac{1}{2} a t^{2} ≃ 2.8 m$

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Masses m1 and m2 are connected by a light string in figure. The wedges is smooth and the pulley is smooth and fixed m1=10kg and m2=7.5kg. When m2 is just released, the distance it will travel in 2 seconds is (g=10m/s2)

The correct option is C 2.8mWe can write two equations for motion of these blocks, m2g−T=m2a T−m1gsin30=m1aPutting the values we get a=2517.5m/s2Now distance travelled by m2 in two seconds = 12at2≃2.8m

Masses $m_{1}$ and $m_{2}$ are connected by a light string in figure. The wedges is smooth and the pulley is smooth and fixed $m_{1} = 10 k g$ and $m_{2} = 7.5 k g$ . When $m_{2}$ is just released, the distance it will travel in 2 seconds is ( $g = 10 m / s^{2}$ )

The correct option is C 2.8m
We can write two equations for motion of these blocks,

$m_{2} g - T = m_{2} a$
$T - m_{1} g s i n 30 = m_{1} a$

Putting the values we get a= $\frac{25}{17.5} m / s^{2}$

Now distance travelled by $m_{2}$ in two seconds = $\frac{1}{2} a t^{2} ≃ 2.8 m$