NEWTON’S SECOND LAW OF MOTION
1.The rate of change of momentum of a body is directly proportional to the external force and the change in momentum takes place in the direction of the force.
2.Newton’s second law of motion leads to a formula useful for measuring force.
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3.Force is a vector. It is always in the direction of change in momentum. Force is also always in the direction of acceleration.
4.SI unit of force is newton (N). If a force acting on a mass of 1 kg produces in it an acceleration of 1 m s-2 in its direction, it is called a newton.
5.CGS unit of force is dyne. If a force acting on a mass of 1 gm produces in it an acceleration of 1 cm s-2 in its direction, it is called a dyne.
6.One newton = 105 dyne.
7.Gravitational units of force: kilogram weight (kg.wt) and gram weight (gm.wt) are called the gravitational units of force. 1 kg.wt or kg. f = 9.8 N, 1 gm.wt or gm.f = 980 dyne.
8.To calculate a force ‘F’, there are several useful variants of the formula .
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Derivation of F = ma : Consider a body of mass ‘m’ moving with initial velocity u. Let
Initial momentum of the body (Pi) = m u
Final momentum of the body (Pf) = m v
Now, change in momentum of the body = Pf – Pi = mv – mu = m (v – u)
Time taken for this change in momentum = (t – 0) = t
Rate of change of momentum = \( \frac{{change\,\,of\,\,momentum}} {{time\,\,taken}} \) = \( \frac{{m\,\,(v\, - \,u)}} {t} \) = m a \( \left( {\because a = \frac{{(v\, - \,u)}} {t}} \right) \)
Change of momentum : If ‘u’ and ‘v’ are the initial and final velocity of a body then its, initial momentum = mu final momentum = mv
Now change of momentum = final momentum – initial momentum = mv – mu
*Change in momentum of a body in different cases
Consider a body of mass m moving with velocity \( \vec V_i \) and momentum \( \vec P_i \). Due to a collision (or) due to the action of a force on it suppose its velocity changes to \( \vec V_i \) and momentum changes to \( \vec P_i \) in a small time interval \( \Delta t \).
Change in momentum of body =\( \Delta \vec P = \vec P_f - \vec P_i \)
\( \left| {\Delta \vec P} \right| = \left| {\vec P_f - \vec P_i } \right| = \sqrt {P_f^2 + P_1^2 - 2P_f P_i \cos \theta } \)
where \( \theta \) = angle between \( \vec P_f \) and \( \vec P_i \)
Case (i) : Consider a body of mass m moving with velocity . If it hits a rigid surface (or) a wall and comes to rest.
Change in momentum of the body = \( \overrightarrow {\Delta P} = \vec P_f - \vec P_i = 0 - \left( {mv} \right)\hat i \)
\( = - \left( {mv} \right)\hat i \)
\( \left| {\overrightarrow {\Delta P} } \right| = mv \)
Note : From law of conservation of linear momentum, theoretically, Change in momentum of surface / wall = +
Case(ii) : In the above case if the body rebounds with same speed V then \( \theta = 180^0 \)
\( \left| {\overrightarrow {\Delta P} } \right| = 2mv \)
Case (iii) : If a body of mass m moving with velocity \( V_1 \hat i \) hits a rigid wall and rebounds with speed V2 then , \( \overrightarrow {\Delta P} = \vec P_f - \vec P_i \)
Momentum of surface / wall = +\( \left( {mv} \right)\hat i \)
Case (iv) : A body of mass m moving with speed V hits a rigid wall at an angle of incidence and reflects with same speed V of body is along the normal, away
from the wall \( \left| {\overrightarrow {\Delta P} } \right| = 2mv\,\,\cos \theta \)
Example-1 Two forces having magnitude 3F and 2F, when act in the same direction simulataneously on a body gives a net force equal to 25 N. Find the value of F.
Solution : Net force = sum of the two forces = 3F + 2F = 5F
\( \begin{gathered} \therefore 5\,F\, = \,25\, \hfill \\ \Rightarrow F = \frac{{25}} {5} = 5N \hfill \\ \end{gathered} \)
Example-2: A car changes its speed from 20 km h-1 to 50km h-1 of mass 3600 kg in 5s. Determine the net external force applied on the car.
Solution: \( F = m\left( {\frac{{v - u}} {t}} \right) = 3600\left( {\frac{{50 - 60}} {5}} \right) \times \frac{{15}} {{18}} \)
= 600 x 30 = 1800 N.
Example-3: If a force of 50 N is applied on a body and it is still at rest then find the magnitude of static fricational force acting on it.
Solution: The magnitude of the force acting on the body (f) = 50 N.
Due to application of this force, the body tends to move but is not set in motion.
and hence, the applied force is equal in magnitude and opposite in direction to static firctional force.
\(\therefore\) The magnitude of static frictional force = 50 N.
Example-4: The speed of a car weighing 1000 kg increases from 36 km/h to 108 km/h. Calculate the change in momentum.
Solution:Mass of the car (m) = 1000 kg
initial velocity (u) = 36 km / h (1 km/h = 5 /18 m/s)
= 36 x 5/ 18 m/s = 10 m/s
Final Velocity (v) = 108 km/h = 108 x 5 / 18 m/s
= 6 x 5
= 30 m/s
Change in momentum = mv -mu
= m(v - u)
= 1000 ( 30 - 10)
= 1000 x 20
=20000
Example-5: An object requries the force of 100 N to gain an acceleration ‘a’. If the mass of the object is 500 kg what will be the value of ‘a’?
Solution: According to the question,
mass (m) = 500 kg,
Force (F) = 100 N,
Acceleration (a) = ?
We know that,Force = Mass x Acceleration
Or F = m x a
Therefore,100 N = 500 kg x a
a = 100 N / 500 kg
a = 100 kg ms-2 / 500 kg
a = 0.2 ms-2
Thus,acceleration of the vehicle = 0.2 ms-2