JEE Main 2015 · 10 Apr · Shift 1 · Physics Q3

← JEE Main 2015 · 10 Apr · Shift 1

PhysicsSingle correct+4 / -1Question 3 of 90

A block of mass m = 10 kg rests on a horizontal table. The coefficient of friction between the block and the table is 0.05. When hit by a bullet of mass 50 g moving with speed $v$ , that gets embedded in it, the block moves and comes to stop after moving a distance of 2 m on the table.

If a freely falling object were to acquire speed $\dfrac v{10}$ after being dropped from height H, then neglecting energy losses and taking $g=10$ ms $^{-2}$ , the value of H is close to :

0.2 km

0.3 km

0.4 km

0.5 km

Answer and worked solution

Official answer: Bonus: full marks to everyone

Friction stops the block: $V^2=2\mu gs=2(0.05)(10)(2)=2$ , so $V=\sqrt2$ m/s. By momentum, $v=\dfrac{10.05}{0.05}\sqrt2\approx284$ m/s.

Then $\frac v{10}=28.4$ m/s, and $H=\dfrac{28.4^2}{20}\approx40$ m $=0.04$ km. No option matches, and the question was given marks to all.

Topics:Elastic and inelastic collisions Friction

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JEE Main 2015 · 10 Apr · Shift 1, Physics question 3