Laws of Motion

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 e

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 $\times$ 15 $\times$ cos 22.5 $°$ = 2.0787 kg-m/s

The final mome

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 r

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

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

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

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 app

(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,

The mass of the man, m = 70 kg, g = 10 m/s2, Let W be the reading on the weighing scale

(a) When the lift is moving upward with uniform speed of 10 m/s, the acceleration = 0

The value of W is given by W = mg = 70 $*10$ = 700 N

(b) The lift is moving downward with acceleration, a = 5m/s² Us

(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,