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V. Numerical Problems · Q2

Q.A particle of mass 5 units is moving with a uniform speed of v=32v=3\sqrt{2} units in the XOY plane along the line y=x+4y=x+4. Find the magnitude of its angular momentum about the origin.

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✓ Free question

Step 1. Recall the formula for a particle moving in a straight line. For a particle of mass mm moving at constant speed vv along a straight path, its angular momentum about any fixed point is L=mvdL=mvd, where dd is the perpendicular (shortest) distance from that point to the line of motion — this stays constant throughout the motion.

Step 2. Find dd, the perpendicular distance from the origin to the line y=x+4y=x+4. Rewrite the line as x−y+4=0x-y+4=0. The perpendicular distance from the origin (0,0)(0,0) to a line ax+by+c=0ax+by+c=0 is ∣c∣a2+b2\dfrac{|c|}{\sqrt{a^2+b^2}}; here a=1,b=−1,c=4a=1,b=-1,c=4:

d=∣4∣12+(−1)2=42=22 units.d=\frac{|4|}{\sqrt{1^2+(-1)^2}}=\frac{4}{\sqrt2}=2\sqrt2\ \text{units}.

Step 3. Substitute into L=mvdL=mvd. Given m=5m=5 units and v=32v=3\sqrt2 units:

L=5×32×22=5×3×2×(2×2)=5×6×2=60.L=5\times3\sqrt2\times2\sqrt2=5\times3\times2\times(\sqrt2\times\sqrt2)=5\times6\times2=60.

Step 4. State the result, noting the units cancel out cleanly because 2×2=2\sqrt2\times\sqrt2=2 exactly, leaving a whole number.

✓Final answer

The magnitude of the angular momentum about the origin is L=60L=60 units.

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