Motion in A Plane Test - 10

Given

\(\vec A= \hat i+\hat j\)

\(\vec B = \hat i - \hat j\)

\(\vec A. \vec B = |A||B|cos\ \theta\)

\(Cos\theta={\vec A.\vec B\over |A||B|}\)

\(= {(\hat i+\hat j).(\hat i-\hat j)\over \sqrt{1^2+1^2}\times \sqrt{1^2+(-1)^2}}={1-1\over2}=0\)

⇒ \(Cos\theta=0\)

\(Cos\theta=Cos90°\)

\(\therefore \ \theta= 90°\). Hence, verifies the option (b).

A scalar quantity does not depend on direction so it does not change for different orientation of axes. So this verifies the option (d).

On resolving the vector, values of cos \(\theta\) or sin \(\theta\) is always less than one, so components have a smaller value.

If \(\vec r\) makes an angle \(\theta\) with x-axis, then components of \(\vec r\) along x-axis = rCos \(\theta\)

It will be maximum if \(Cos\theta_{max}\) = 1 ⇒ \(\theta\) = 0°

Hence, given vector \(\vec r\) is along positive X-axis.

a projectile is fired at \(\theta\) = 15°, R=50 m

\(R={u^2sin2\theta\over g}\)

\(50={u^2sin2\times15°\over g}\) ⇒ \(u^2=50g\times 2\)

\(u^2=100g\)

Now \(\theta\) = 45°, \(u^2=100g\)

\(\therefore\) \(R={u^2sin2\theta\over g}\) = \(100g\times sin2\times45°\over g\)

⇒ R = 100 m

So, this verifies option (c)

In the two-dimensional motion, If instantaneous speed is constant then equal path lengths are traversed in equal intervals of time because speed is a scalar quantity.

We know that change in acceleration and velocity is in the direction of Force (F) by \(\vec F = m \vec a\) and change in velocity is zero so acceleration will also be zero and will be in the same planes as that of velocity.

(\(\vec A\) x \(\vec B\)) = \(\vec X\)

The direction of x is perpendicular to both planes containing A and B.

(\(\vec A\) x \(\vec B\)) x \(\vec C\) = \(\vec X\) x \(\vec C\) = \(\vec Y\)

The direction of \(\vec Y\) is perpendicular to the plane of \(\vec X\) and \(\vec C\) which again become in the plane of \(\vec A\),\(\vec B\),\(\vec C\) but perpendicular to the plane of \(\vec X\) and \(\vec C\) Hence, option (c) is verified.

\(|\vec A+\vec B|^2=|\vec A|^2\)

\(|\vec A|^2+|\vec B|^2+2|\vec A||\vec B|cos\theta=|\vec A|^2\)

\(|\vec B|^2+2|\vec A||\vec B|cos\theta=0\)

\(|\vec B|\ \ \ [|\vec B|+2|\vec A|cos\theta]=0\)

\(|\vec B|=0\)

verifies option (a).

Or \(|\vec B|+2|\vec A|cos\theta=0\)

\(cos\theta={-|\vec B|\over 2|\vec A|}\)

Maximum height H of projectile H = \(u^2sin^2\theta\over 2g\)

from here \(H\propto sin^2\theta\)

For Projectile 1, \(\theta=\theta_1\)

For Projectile 2, \(\theta=\theta_2\)

\(u_1=v_o\,\,\,\,\,u_2=v_0\)

(a) \(H_1>H_2\)     (Given)

\({v^2_0sin^2\theta_1\over 2g}>{v^2_0sin^2\theta_2\over 2g}\)

\(sin^2\theta_1>sin^2\theta_2\)

\(sin^2\theta_1-sin^2\theta_2>0\)

\((sin\theta_1-sin\theta_2)(sin\theta_1+sin\theta_2)>0\)

thus either \((sin\theta_1-sin\theta_2)>0\) 

\(\Rightarrow\, sin\theta_1 >sin\theta_2\)

or \((sin\theta_1+sin\theta_2)>0\)

⇒ \(\theta_1\) and \(\theta_2\) lies between     \(sin\theta_1>sin\theta_2\)

0° to 90° i.e., acute as given \(\theta_1\) > \(\theta_2\)    ....(i)

so option (a) is correct

(b) we know time of flight

\(T={2u\,sin\theta\over g}\)

\(\therefore\) \(T_1={2v_0\,sin\theta_1\over g}\) and \(T_2={2v_0\,sin\theta_2\over g}\)

\({T_1\over T_2}={sin\theta_1\over sin\theta_2}\) ⇒ \(T_1sin\theta_2=T_2sin\theta_1\)

As sin\(\theta_1\) > sin\(\theta_2\)

\(\therefore\) \(T_1>T_2\)

So, Option (b) is also correct.

For option (a):

(i) If acceleration is uniform then

\(\vec \upsilon_{av}={\vec u+\vec v\over2}\) or \(\vec \upsilon_{av}={1\over2}[\vec \upsilon(t_2)-\vec \upsilon(t_1)]\) is correct only when acceleration is uniform, but it not given.

So option (a) is verified that given relation is incorrect.

For option (c): In given relation displacement is half of rate of change of velocity which is not possible because dimensions in LHS is \([M^0L^ 1T ^0 ]\) but in RHS. \([M^0L^ 1T ^{-2} ]\) which are not equal. So relation is incorrect. 

(i) For option (a): In uniform circular motion speed is always constant, so verified option(a).

(ii) For option (b): The direction of velocity is always tangentially and towards motion, if a string breaks up the particle moves tangentially. So verifies the option (b).

(iii) For option (c): The direction of acceleration \(({v^2\over r})\) is always along the direction of force applied by Newton’s IInd law of motion. The force is applied on the string to rotate the particle in circular motion, hence the direction of acceleration is along the string towards the center of circular path. As string is moving in circular plane with particle so the direction of acceleration changes always and verifies the option (c).

\(|\vec A+\vec B|\) = \(|\vec A-\vec B|\)   (given)

\(|\vec A+\vec B|^2\) = \(|\vec A-\vec B|^2\)    (squaring both sides)

\(|\vec A|^2+|\vec B|^2+2|\vec A||\vec B|cos\theta=|\vec A|^2+|\vec B|^2-2|\vec A||\vec B|cos\theta\)

\(4|\vec A||\vec B|cos\theta=0\)

⇒ \(4\ne0, |\vec A|=0\) or \(|\vec B|=0\) or cos \(\theta\) = 0

⇒ cos \(\theta\) = cos 90°

⇒ \(\theta\) = 90°

⇒ |A| = |B| = 0

So, It does not verify the option (a) and (c) and verifies the option (b) and (d).