A particle moves in the x - y plane under the influence of a force such that its linear momentum is pt -A-coskt-sinkt- where A and k are constants- The angle between the force and the momentum is

P⃗(t) =A[îcos(kt) ĵsin(kt)] ⇒F⃗(t) = dp⃗(t)/dt = Ak [ î sin(kt) ĵcos(kt)] Now, finding the dot product, p⃗(t) . F⃗(t) = 0 Hence, angle is 90^∘. ...

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Standard XI

Physics

COM in Collision and Explosion

A particle mo...

Question

A particle moves in the x – y plane under the influence of a force such that its linear momentum is $p (t) = A [^i c o s (k t) -^j s i n (k t)]$ , where A and k are constants. The angle between the force and the momentum is

$0^{\circ}$

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$30^{\circ}$

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$45^{\circ}$

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$90^{\circ}$

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Solution

The correct option is D
$90^{\circ}$

$\to p (t) = A [^i c o s (k t) -^j s i n (k t)] \Rightarrow \to F (t) = \frac{d \to p (t)}{d t} = A k [-^i s i n (k t) -^j c o s (k t)]$
Now, finding the dot product, $\to p (t) . \to F (t) = 0$
Hence, angle is $90^{\circ}$ .

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A particle moves in the x – y plane under the influence of a force such that its linear momentum is p(t) =A[^icos(kt)−^jsin(kt)], where A and k are constants. The angle between the force and the momentum is

The correct option is D 90∘ →p(t) =A[^icos(kt)−^jsin(kt)]⇒→F(t)=d→p(t)dt=Ak[−^isin(kt)−^jcos(kt)] Now, finding the dot product, →p(t).→F(t)=0 Hence, angle is 90∘.

A particle moves in the x – y plane under the influence of a force such that its linear momentum is $p (t) = A [^i c o s (k t) -^j s i n (k t)]$ , where A and k are constants. The angle between the force and the momentum is

The correct option is D
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$\to p (t) = A [^i c o s (k t) -^j s i n (k t)] \Rightarrow \to F (t) = \frac{d \to p (t)}{d t} = A k [-^i s i n (k t) -^j c o s (k t)]$
Now, finding the dot product, $\to p (t) . \to F (t) = 0$
Hence, angle is $90^{\circ}$ .