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Standard XII

Physics

Impulsive Forces in Collision

A jet of wate...

Question

A jet of water with cross section of $6 {c m}^{2}$ strikes a wall at an angle of $60^{\circ}$ to the normal and rebounds elastically from the wall without losing energy. If the velocity of the water in the jet is $12 m / s$ , the force acting on the wall is

0.864 N

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86.4 N

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72 N

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7.2 N

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Solution

The correct option is B $86.4 N$
F = $\frac{d P}{d t}$
= $\frac{2 d m V cos 60^{o}}{d t}$
= $\frac{2 (ρ A d x) V cos 60^{o}}{d t}$
= $2 ρ A V^{2} cos 60^{o}$
= $10^{3} \times 6 \times 10^{- 4} \times 12^{2}$
= $86.4 N$

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A jet of water with cross section of 6cm2 strikes a wall at an angle of 60∘ to the normal and rebounds elastically from the wall without losing energy. If the velocity of the water in the jet is 12m/s, the force acting on the wall is

The correct option is B 86.4NF = dPdt = 2dmVcos60odt = 2(ρAdx)Vcos60odt = 2ρAV2cos60o = 103×6×10−4×122 =86.4N

A jet of water with cross section of $6 {c m}^{2}$ strikes a wall at an angle of $60^{\circ}$ to the normal and rebounds elastically from the wall without losing energy. If the velocity of the water in the jet is $12 m / s$ , the force acting on the wall is

The correct option is B $86.4 N$
F = $\frac{d P}{d t}$
= $\frac{2 d m V cos 60^{o}}{d t}$
= $\frac{2 (ρ A d x) V cos 60^{o}}{d t}$
= $2 ρ A V^{2} cos 60^{o}$
= $10^{3} \times 6 \times 10^{- 4} \times 12^{2}$
= $86.4 N$