Laws of Motion Test - 15

In the given system,

a = \(\frac{(m_1 - m_2) g}{m_1 + m_2} = \frac{g}{8}\)

\(\therefore \frac{m_1 - m_2}{m_1 + m_2} = \frac{1}{8}\)

\(8m_1 - 8m_2 = m_1 + m_2\)

\(7m_1 = 9m_2\)

\(\frac{m_1}{m_2} = \frac{9}{7}\)

Solve two equations

\(R^2 = 9P^2 + 4P^2 + 12P^2 cos \theta\) \(= 13P^2 + 12P^2cos \theta \) .....(1)

\((2R^2) = (2 \times 3P)^2 + 4P^2 + 24P^2 cos \theta\) = \(40P^2 + 24P^2 cos \theta\) ....(2)

Multiplying (1) by 4, \(4R^2 = 52P^2 + 48P^2 cos \theta \)  .....(3)

Equating (2) and (3)

\(40 P^2 + 24P^2 cos \theta\) \( = 52P^2 + 48P^2 cos \theta\)

⇒ \(12P^2 = -24P^2 cos \theta\)

⇒ \(cos \theta = -\frac{1}{2} = cos 120^o\)

Let be vector p and q then

|p - q| \(\leq\) |r| \(\leq\) |p + q|

using this concept

take p = 5 and q = 10 so

|5 - 10| \(\leq\) |r| \(\leq\) |5 + 10|

|-5|\(\leq\) |r| \(\leq\) |15|

5 \(\leq\) |r| \(\leq\) 15

So range should between 5 to 15 and in this range 4 is not present.

Tension, thrust, weight are all common forces in mechanics whereas impulse is not a force.

Impulse = force × Time duration

\(F = \frac{d}{dt}(mv) = v \frac{dm}{dt}\)

⇒ 210 = 300 x \(\frac{dm}{dt}\)

⇒ \(\frac{dm}{dt} = 0.7 kg/s\)

If a large force F acts for a short time dt the impulse imparted I is

I = Fdt = \(\frac{dp}{dt}dt,\) 

I = dp = change in momentum

Lift accelerating down with 'a'

a < g

mg - R = ma

R = m(g - a)

Apparent weight < Actual weight

For the man standing in the lift, the acceleration of the ball.

\(\overrightarrow {a_{bm}} = \overrightarrow {a_b} - \overrightarrow {a_m}\) ⇒ \(\overrightarrow {a_{bm}} = g - a\)

For the man standing on the ground, the acceleration of the ball.

\(\overrightarrow {a_{bm}} = \overrightarrow {a_b} - \overrightarrow {a_m}\) ⇒ \(\overrightarrow {a_{bm}} = g - 0 = g\)

Fdt = mv - mu from work energy theorem

\(F = \frac{m(v-u)}{dt} = \frac{5(65 - 15)}{0.2 \times 100}\)

= \(\frac{5 \times 50}{20}\) = 12.5 N

When the lift moves upwards, the apparent weight,[= m(g + a)]. Hence reading of spring balance increases.

Swimming is account on Newton’s third law of motion since as we push the water we feel equal force by water on us and we move front.

(i) Very high since the force depends upon the change in momentum and since the change of momentum of bird before and after collision is very high.

(ii) Therefore the force experienced by the bird is very high with respect to the plane.

For the equilibrium of force, the resultant of two smaller is equal and opposite of third one.

\(C^2 = A^2 + B^2\)

\(5^2 = 3^2 + 4^2\)

25 = 25

Momentum = mass x velocity

Momentum = 0.05 x 400 = 20kg. \(\frac{m}{sec}\)

Accelerating force on the rocket = 20 N

When a bicycle is in motion, two cases may arise:

(i) When the bicycle is being pedaled. In this case, the applied force has been communicated to rear wheel. Due to which the rear wheel pushes the earth backward. Now the force of friction acts in the forward direction on the rear wheel but front-wheel move forward due to inertia, so force of friction works on it in backward direction

(ii) When the bicycle is not being pedaled: In this case both the wheels move in forward direction, due to inertia. Hence force of friction on both the wheels acts in backward direction.

Let distance travel in \(1^{st}\) second, \(2^{nd}\) second, \(3^{rd}\) be \(\mathrm{S_1,S_2\,\,and\,\,S_3}\) respectively.

Applying equation of motion,

For, t = 1s

\(\mathrm{S_1=u\times t_1+\frac{1}{2at^2_1}}\)

\(\mathrm{5=u\times1+\frac{1}{2}\times a\times1^2}\)

\(\mathrm{10=2u+a}\)  ......(1)

For, t  = 2s

\(\mathrm{S_1+S_2=u\times t+\frac{1}{2at^2}}\)

\(\mathrm{S_1+S_2=u\times 2+\frac{1}{2}\times a\times2^2}\)

\(\mathrm{S_1+S_2=2u+2a}\) ....(2)

For, t = 3s

\(\mathrm{S_1+S_2+S_3=u\times t^2+\frac{1}{2at^2}}\)

\(\mathrm{S_1+S_2+2=u\times3+\frac{1}{2a}\times9}\)  ...(3)

Substituting value of \(\mathrm{S_1+S_2}\) from eqn 2 and solving,

we get, \(\mathrm{a=-1.5m/s^2}\)

\(\mathrm{F=ma}\)

\(\mathrm{F=4\times(-1.5)}\)

\(\mathrm{F=-6N}\)