JEE MAIN - Physics (2021 - 18th March Morning Shift - No. 22)
A ball of mass 10 kg moving with a velocity 10$$\sqrt 3 $$ m/s along the x-axis, hits another ball of mass 20 kg which is at rest. After the collision, first ball comes to rest while the second ball disintegrates into two equal pieces. One piece starts moving along y-axis with a speed of 10 m/s. The second piece starts moving at an angle of 30$$^\circ$$ with respect to the x-axis. The velocity of the ball moving at 30$$^\circ$$ with x-axis is x m/s. The configuration of pieces after collision is shown in the figure below. The value of x to the nearest integer is ____________.
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Answer
20
Explanation
Velocity of 10 kg ball = $${v_{10}} = 10\sqrt {3\widehat i} $$
initial total momentum of system = $$10 \times 10\sqrt {3\widehat i} $$
Final total momentum of system
$$ = 10 \times 10\widehat j + 10 \times x(\cos 30^\circ \widehat i - \sin 30^\circ \widehat j)$$
Now by conservation of momentum
$$10 \times 10\sqrt {3\widehat i} = 10 \times 10\widehat j + 10 \times x\left( {{{\sqrt 3 } \over 2}\widehat i - {1 \over 2}\widehat j} \right)$$
$$ \Rightarrow $$ x = 20
initial total momentum of system = $$10 \times 10\sqrt {3\widehat i} $$
Final total momentum of system
$$ = 10 \times 10\widehat j + 10 \times x(\cos 30^\circ \widehat i - \sin 30^\circ \widehat j)$$
Now by conservation of momentum
$$10 \times 10\sqrt {3\widehat i} = 10 \times 10\widehat j + 10 \times x\left( {{{\sqrt 3 } \over 2}\widehat i - {1 \over 2}\widehat j} \right)$$
$$ \Rightarrow $$ x = 20
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