JEE MAIN - Physics (2019 - 11th January Morning Slot - No. 9)
A body of mass 1 kg falls freely from a height of 100 m, on a platform mass 3 kg which is mounted on a spring having spring constant k = 1.25 $$ \times $$ 106 N/m. The body sticks to the platform and the spring's maximum compression is found to be x. Given that g = 10 ms–2
, the value of x will be close to :
8 cm
4 cm
40 cm
80 cm
Explanation
velocity of 1 kg block just before it collides with 3kg block
= $$\sqrt {2gh} = \sqrt {2000} $$ m/s
Applying momentum conversation just before and just after collision.
1 $$ \times $$ $$\sqrt {2000} $$ = 4v $$ \Rightarrow $$ v = $${{\sqrt {2000} } \over 4}$$ m/s
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initial compression of spring
1.25 $$ \times $$ 106 x0 = 30 $$ \Rightarrow $$ x0 $$ \approx $$ 0
applying work energy theorem,
Wg + Wsp = $$\Delta $$KE
$$ \Rightarrow $$ 40 $$ \times $$ x + $${1 \over 2}$$ $$ \times $$ 1.25 $$ \times $$ 106 (02 $$-$$ x2)
= 0 $$-$$ $${1 \over 2}$$ $$ \times $$ 4 $$ \times $$ v2
solving x $$ \approx $$ 4 cm
= $$\sqrt {2gh} = \sqrt {2000} $$ m/s
Applying momentum conversation just before and just after collision.
1 $$ \times $$ $$\sqrt {2000} $$ = 4v $$ \Rightarrow $$ v = $${{\sqrt {2000} } \over 4}$$ m/s
initial compression of spring
1.25 $$ \times $$ 106 x0 = 30 $$ \Rightarrow $$ x0 $$ \approx $$ 0
applying work energy theorem,
Wg + Wsp = $$\Delta $$KE
$$ \Rightarrow $$ 40 $$ \times $$ x + $${1 \over 2}$$ $$ \times $$ 1.25 $$ \times $$ 106 (02 $$-$$ x2)
= 0 $$-$$ $${1 \over 2}$$ $$ \times $$ 4 $$ \times $$ v2
solving x $$ \approx $$ 4 cm
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