A particle having a mass 0.5 kg is projected under gravity with a speed of 98 m/sec at an angle of 60º. The magnitude of the change in momentum (in N-sec) of the particle after 10 seconds is

Solution

There is no change in horizontal velocity,hence no change in momentum in horizontal direction. The vertical velocity at t = 10sec is

v=98 × sin 60º-(9.8) × 10 = –13.13 m/sec

so change in momentum in vertical direction is

= (0.5 × 98 × √3/2)-[-(0.5 × 13.13)]

= 42.434 + 6.56 = 48.997 ≈ 49

A projectile is moving at 60 m/s at its highest point, where it breaks into two equal parts due to an internal explosion.One part moves vertically up at 50 m/s with respect to the ground. The other part will move at a speed of

110 m/s
120 m/s
130 m/s
10\(\sqrt{61}\) m/ s

Solution

A body is projected at an angle of 30º to the horizontal with speed 30 m/s. What is the angle with the horizontal after 1.5 seconds? Take g= 10 m/s².

0º
30º
60º
90º

Solution

u_x=30 cos 30º=30√3/2,,u_y= 30 sin 30°,

v_y=30 sin 30º - gt

v_y=30 sin 30º - 10 × 1.5=0

As vertical velocity = 0,Angle with horizontal α = 0º

It is a state, when a particle reach to a highest point of its path.

A cannon on a level plane is aimed at an angle θ above the horizontal and a shell is fired with a muzzle velocity v₀ towards a vertical cliff a distance D away. Then the height from the bottom at which the shell strikes the side walls of the cliff is

\(D\, sin\theta -\frac{gD^{2}}{2V_{0}^{2}sin^{2}\theta }\)
\(D\, cos\theta -\frac{gD^{2}}{2V_{0}^{2}cos^{2}\theta }\)
\(D\, tan\theta -\frac{gD^{2}}{2V_{0}^{2}tan^{2}\theta }\)
\(D\, tan\theta -\frac{gD^{2}}{2V_{0}^{2}sin^{2}\theta }\)

Solution

From the resultant path of the particle, when it is projected at angle θ with its velocity u is

A force of –\(F\hat{k}\) acts on O,the origin of the coordinate system. The torque about the point (1, –1) is

\(F(\hat{i}-\hat{j})\)
\(F(\hat{i}-\hat{j})\)
\(F(\hat{i}+\hat{j})\)
\(-F(\hat{i}-\hat{j})\)

Solution

At what angle must the two forces (x + y) and (x – y) act so that the resultant may be \(\sqrt{(x^{2}+y^{2})}\)?

cos^-1[-(x²+y²)/2(x²-y²)]
cos^-1[-2(x²-y²)/2(x²+y²)]
cos^-1[-(x²+y²)/2(x²-y²)]
cos^-1[-(x²-y²)/2(x²+y²)]

Solution

Let \(\overrightarrow{C}=\overrightarrow{A}+\overrightarrow{B}\)

\(\left |\overrightarrow{C} \right |\) is always greater than \(\left |\overrightarrow{A} \right |\)
It is possible to have \(\left |\overrightarrow{C} \right | < \left | \overrightarrow{A} \right | and \left |\overrightarrow{C} \right | < \left | \overrightarrow{B} \right |\)
\(\overrightarrow{C}\) is always equal to \(\overrightarrow{A}+\overrightarrow{B}\)
\(\overrightarrow{C}\) is never equal to \(\overrightarrow{A}+\overrightarrow{B}\)

The resultant of two forces 3P and 2P is R. If the first force is doubled then the resultant is also doubled. The angle between the two forces is

60º
120º
70º
180º

Solution

Solve two equations : R² = 9P²+ 4P²+ 12P² cosθ and 4R²= 36P²+ 4P²+ 24P²cosθ.

The angles which the vector \(\overrightarrow{A}=3\hat{i}+6\hat{j}+2\hat{k}\) makes with the co-ordinate axes are :

cos^-1 ³⁄₇,cos^-1 ⁴⁄₇,cos^-1¹⁄₇
cos^-1 ³⁄₇,cos^-1 ⁶⁄₇,cos^-1²⁄₇
cos^-1 ⁴⁄₇,cos^-1 ⁵⁄₇,cos^-1³⁄₇
None of these

Solution

For any two vectors \(\overrightarrow{A}\) and \(\overrightarrow{A}\), if \(\overrightarrow{A}.\overrightarrow{B}=\left |\overrightarrow{A}\times \overrightarrow{B} \right |\), the magnitude of \(\overrightarrow{C}=\overrightarrow{A}+\overrightarrow{B}\) is

A + B
\(\sqrt{A^{2}+B^{2}+\sqrt{2}AB}\)
\(\sqrt{A^{2}+B^{2}}\)
\(\sqrt{A^{2}+B^{2}+\frac{AB}{\sqrt{2}}}\)

Solution

UnAttempted

0

Attempted

0

Correct

0

Wrong

0

Complete