Laws of Motion

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

Mass of the body A, m₁ = 10 kg, mass of the body B, m₂ = 20 kg, Horizontal force = 600 N

Total mass of the system, m =m₁ + m₂ = 30 kg

From the relation F = ma, we get a = 600 / 30 = 20 m/s²

(i) When the force is applied on A

F – T = m₁a or T = F - m₁a = 600 – 10 $*20$ = 400 N

(ii) When the force is applied on B

F-T = m₂ a, T = F - m₂ a = 600 – 20 $*20=200$ N

(a) For t<0, the distance covered by the particle x is zero. Hence force on the particle is zero. For t>4s, the particle is moving at a constant distance, so the force will be zero. For 0

(b) Impulse is given by the equation, Impulse = total change of momentum. At t = 0, u = 0, v = distance / time = ¾ = 0.75 m/s

Impulse = 4 $*$  (0.75-0) kg-m/s = 3 kg-m/s

Impulse at t=4, u = 0.75, v = 0, Impulse = 4   (0-0.75) = -3 kgm/s

(a) When the bob is at extreme position, the velocity at that point is zero. If the string is cut at that location, the bob will fall because of its own weight

(b) At the mean position, if the string is cut, due to horizontal velocity, the bob will act like a projectile and will fall on the ground, following a parabolic path.

The acceleration of the truck, a = 2 m/s², t = 10 s, Initial velocity, u = 0, from the equation v = u + at, we get v = 2 $*10$ = 20 m/s

At time t = 11 s, the horizontal component of the velocity Vx remains unchanged due to no air resistance. Hence Vx = 20 m/s

The vertical component of the velocity Vy is expressed as

Vy = u + $a_y$ t

Here t =11-10 = 1 s,  $a_y=a=g=$ 10 m/s², u = 0

Vy = 10 m/s

The resultant velocity V is given by V = ( $V_x^2$ + $V_y^2$ )1/2 = 22.36 m/s

$\tan ??$ =Vy/Vx=10/20,  $?=$ 26.57 $°$ w.r.t. horizontal