Class 11th

The acceleration of the conveyer belt, a = 1 m/s²

Coefficient of static friction,  ${\mu }_s$ = 0.2

Mass of the man, m = 65 kg

The net force experienced by the man, MA = 1 $*65$ N = 65 N

This net force is due to the friction between the belt and the man

At maximum static friction

${\mu }_s*mg=m{a}_{max}$

$a_{max}$ = ${\mu }_s*g$ = 0.2 $*10=2$ m/s²

This graph could be of a ball rebounding between two walls separated by a distance of 2 cm.

The ball rebounds every 2 secs between the walls with uniform velocity.

Velocity of rebounding = displacement / time = ( 2 $*{10}^{-2}$ )/2 = 0.01 m/s

Initial momentum = mu = 0.04 $*0.01=$ 4 $*{10}^{-4}$ kgm/s

Final momentum = -mu = -4 $*{10}^{-4}$ kgm/s

Magnitude of Impulse = Initial momentum – final momentum = 8 $*{10}^{-4}$ kgm/s

The time between two consecutive impulses is 2 secs, so the ball receives an impulse every 2 seconds.

(a) & (c) solution not possible as the stone will fly off tangentially. The answer is (b)

Mass of the stone, m = 0.25 kg

Radius of the circular path = 1.5 m

Speed = 40 rev/min = 0.67 rev/s

Angular velocity,  $?$ = 2 $?n$ = 2 $*\frac{22}{7}*0.67$ = 4.19 rev/s

The tension of the string, T = m ${?}^2$ r = 0.25 $*{4.19}^2*1.5$ = 6.59 N

Maximum tension of the string,  $T_{max}$ = 200 N

From the equation $T_{max}$ = ${mv}_{max}^2$ /r,  $v_{max}^2$ = $T_{max}*r/m$

$v_{max}^2$ = 200 $*1.5$ /0.25

$v_{max}$ = 34.64 m/s

Velocity of the ball, u = 54 km/h = 15 m/s, mass of the ball, m = 0.15 kg

The ball is deflected back such that the included angle is 45 $°$

The initial momentum of the ball, along the direction of NO is = mu $\cos ?\theta$ = 0.15 $*$ 15 $*$ cos 22.5 $°$ = 2.0787 kg-m/s

The final momentum of the ball = mu $\cos ?\theta$ along ON

Impulse = change of momentum = mu $\cos ?\theta$ - (-mu $\cos ?\theta$ ) = 2mu $\cos ?\theta$ = 4.16 kg-m/s

The mass of the shell, m = 0.020 kg

The mass of the gun, M = 100 kg

Speed of the shell, v = 80 m/s

The initial velocity of the shell and the gun = 0, so the initial momentum of the system = 0

Applying the law of conservation of momentum, the initial momentum = final momentum

0 = mv – MV, where V is the recoil speed of the gun

V = mv/M = 0.020 $*$ 80 / 100 = 0.016 m/s

Mass of each ball, m = 0.05 kg

Initial velocity of each ball, u = 6 m/s

The initial momentum of each ball before collision = mu = 0.05 x 6 = 0.3 kg-m/s

After the collision, the ball changes the direction of motion without the change in magnitude of the velocity, so

The final momentum after collision of the first ball = - 0.05 $*$ 6 = - 0.3 kg-m/s

The final momentum after collision of the second ball = 0.05 $*$ 6 = 0.3 kg-m/s

Impulse imparted to the first ball = (-0.3) – (0.3) = -0.6 kg-m/s

Impulse imparted to the second ball = (0.3) – (-0.3) = 0.6 kg-m/s

The two impulses are opposite in direction.

m = m₁ + m₂

If $v_1$ and $v_2$ be the corresponding velocities of two nuclei then total linear momentum after disintegration = m₁ $v_1$ + m₂ $v_2$

Since at the initial stage, mass nuclei was at rest, so the initial linear momentum = 0

From the law of conservation we know

Total linear momentum before disintegration = total momentum after disintegration

0 = m₁$v_1$ + m₂$v_2$

$v_1$ = - m₂$v_2$ / m₁

-ve sign indicates that the two velocities $v_1$ and $v_2$ are in opposite direction

Given, small mass, m₁ = 8 kg. bigger mass, m₂ = 12 kg

Let T be the tension in the string

The heavier mass will go downward and the lighter mass will go upward.

Applying Newton's 2^nd law,

For mass m_1, T –m₁g = m₁a …… (1)

For mass m₂, m₂g – T = m₂a……. (2)

Adding (1) & (2), we get

(m_{2 -} m₁₎ g _{= (}m_{1 +} m₂₎ a

a = (m_{2 -} m₁₎ g/ ₍m_{1 +} m_{2) =} 4 $*10/20$ = 2 m/s

From equation (1), we get, T = 8 (2 +10) = 96 N