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UPSEE 2019 Solved Paper

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Question : 5 of 150

Marks: +1, -0

The linear momentum of a particle moving in X-Y plane under the influence of a force is given as

P(t)=A(\hat{i} \cos bt - \hat{j} \sin bt)

A

and b are constants. The angle between the force and momentum is

$0^{\circ}$
$45^{\circ}$
$60^{\circ}$
$90^{\circ}$

Solution:

Here, linear momentum of a particle

p=A(\hat{i} \cos bt - \hat{j} \sin bt)

.......(i)

where,

A

and

B

are constants.

As we know that,

F=\frac{dp}{dt}

From Eq. (i), we get

\Rightarrow F=\frac{d}{dt}(A \hat{i} \cos b t - A \hat{j} \sin b t)

\Rightarrow F=-A \hat{i} b \sin b t - A \hat{j} b \cos b t

Now, dot product

=F \cdot p

=(-A \hat{i} b \sin b t - A \hat{j} b \cos b t) \cdot

(A \hat{i} \cos b t - A \hat{j} \sin b t)

=-A b \sin b t \cos b t + A b \sin b t \cos b t

=0

Since,

F \cdot p = F p \cos \theta = 0

\Rightarrow \cos \theta = 0 \Rightarrow \theta = 90^{\circ}

Hence, the angle between momentum

p

and force F is

90^{\circ}

.

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