$A$ truck of mass $M$ and a car of mass $\frac{M}{10}$ moving with the same momentum are brought to a halt by the application of the same braking force. The ratio of the distances travelled by the truck and the car before they come to a stop is

$A$ slide with a frictionless curved surface,which becomes horizontal at its lower end,is fixed on the terrace of a building of height $3h$ from the ground,as shown in the figure. $A$ spherical ball of mass $m$ is released on the slide from rest at a height $h$ from the top of the terrace. The ball leaves the slide with a velocity $\vec{u}_0 = u_0 \hat{x}$ and falls on the ground at a distance $d$ from the building,making an angle $\theta$ with the horizontal. It bounces off with a velocity $\vec{v}$ and reaches a maximum height $h_1$. The acceleration due to gravity is $g$ and the coefficient of restitution of the ground is $e = 1 / \sqrt{3}$. Which of the following statement$(s)$ is(are) correct?
$(A)$ $\vec{u}_0 = \sqrt{2gh} \hat{x}$
$(B)$ $\vec{v} = \sqrt{2gh} \hat{x} + \sqrt{2gh} \hat{z}$
$(C)$ $\theta = 60^{\circ}$
$(D)$ $d / h_1 = 2\sqrt{3}$

$A$ block of mass $1 \ kg$ is placed at point $A$ on a rough path. It is gently pushed to the right. It slides down the slope and reaches point $B$. Find the work done by the friction force on the block during the journey from point $A$ to point $B$ in $J$. (Assume the vertical height difference between $A$ and $B$ is $0.2 \ m$ and the block starts and ends at rest).

$A$ bullet of mass $m_1$ is moving with speed $v_0$ and hits a sand bag of mass $m_2$. If the speed of the bullet after passing through the sand bag is $\frac{v_0}{3}$,then the height $h$ up to which the bag rises is (assume,$g=$ acceleration due to gravity).

$A$ particle of mass $m$ is moving along the $x$-axis with initial velocity $u \hat{i}$. It collides elastically with a particle of mass $10m$ at rest and then moves with half its initial kinetic energy (see figure). If $\sin \theta_{1} = \sqrt{n} \sin \theta_{2}$,then the value of $n$ is:

$A$ body of mass $4 \ kg$ is moving with a momentum of $8 \ kg \ m/s$. $A$ force of $0.2 \ N$ acts on it in the direction of motion for $10 \ s$. The increase in kinetic energy in joules is: