A body is placed on a rough inclined plane of inclination $\mathrm{\theta }$. As the angle $\mathrm{\theta }$ is increased from 0^o to 90^o , the contact force between the block and the plane

1. Remains constant

2. First remains constant then decreases

3. First decreases then increases

4. First increases then decreases

Internal forces can change:

1. the linear momentum but not the kinetic energy of the system.

2. the kinetic energy but not the linear momentum of the system.

3. linear momentum as well as kinetic energy of the system.

4. neither the linear momentum nor the kinetic energy of the system.

In a cricket match, the fielder draws his hands backward after receiving the ball in order to take a catch because:

1. His hands will be saved from getting hurt.

2. He deceives the player.

3. It is a fashion.

4. He catches the ball firmly.

A ball is traveling with uniform translatory motion. This means that:

A meter scale is moving with uniform velocity. This implies:

A cricket ball of mass 150 g has an initial velocity $\mathrm{u}=(3\hat{\mathrm{i}}+4\hat{\mathrm{j}}){\mathrm{ms}}^{-1}$ and a  final velocity$\mathrm{v}=-(3\hat{\mathrm{i}}+4\hat{\mathrm{j}}){\mathrm{ms}}^{-1}$, after being hit. The change in momentum (final momentum-initial momentum) is (in kgm/s)

\(1.~zero\)
\(2.~-\left ( 0.45\hat{i}+0.6\hat{j} \right ) \)
\(3.~-\left ( 0.9\hat{i}+1.2\hat{j} \right ) \)
\(4.~-5\left ( \hat{i} +\hat{j}\right ) \)

Conservation of momentum in a collision between particles can be understood from:

A hockey player is moving northward and suddenly turns westward at the same speed to avoid an opponent. The force that acts on the player is

1. frictional force along westward
2. muscle force along southward
3. frictional force along south-West
4. muscle force a south-West

A body of mass 2kg travels according to the law $\mathrm{x}\left(\mathrm{t}\right)=\mathrm{pt}+{\mathrm{qt}}^2+{\mathrm{rt}}^3$ where, $\mathrm{q}=4{\mathrm{ms}}^{-2},\mathrm{p}=3{\mathrm{ms}}^{-1}\text{ and }\mathrm{r}=5{\mathrm{ms}}^{-3}$. The force acting on the body at t=2s is

1. 136 N
2. 134 N
3. 158 N
4. 68 N

A body with a mass of \(5\) kg is acted upon by a force \(\vec{F}=\left ( -3\hat{i} +4\hat{j}\right )\) N. If its initial velocity at \(t=0\) is \(\vec{v}=\left ( 6\hat{i} -12\hat{j}\right )\) m/s, the time at which it will just have a velocity along the Y-axis is:
1. never
2. \(10\) s
3. \(2\) s
4. \(15\) s