LEARNATHON

III

Competition for grade 6 to 10 students! Learn, solve tests and earn prizes!

Learn more### Theory:

**Newton’s second law of motion**:

According to Newton’s second law of motion, “the force acting on an object is directly proportional to the rate of change of linear momentum of the object, and the change in momentum takes place in the direction of the force”.

This law aids us to determine the amount of force. So, it is also known as the ‘law of force’.

Consider,

‘\(m\)’ be the mass of a moving object, moving along a straight line with an initial speed of ‘\(u\)’ after the time ‘\(t\)’, the velocity of the object changes to ‘\(v\)’ due to the influence of an unbalanced external force \(F\).

$\begin{array}{l}\mathit{Initial}\mathit{momentum}\mathit{of}\mathit{the}\mathit{body}({P}_{i})\; =m\times u\\ \\ \mathit{Final}\mathit{momentum}\mathit{of}\mathit{the}\mathit{body}({P}_{f})\phantom{\rule{0.147em}{0ex}}=m\times v\\ \\ \mathit{Change}\mathit{in}\mathit{momentum}(\mathrm{\Delta}p)\; =\phantom{\rule{0.147em}{0ex}}{P}_{f}-{P}_{i}\\ \\ \mathrm{\Delta}p\phantom{\rule{0.147em}{0ex}}=\; (m\times v)\; \u2013\; (m\times u)\\ \\ \mathit{By}\mathit{Newton}\u2019s\mathit{second}\mathit{law}\mathit{of}\mathit{motion},\\ \\ \mathit{Force}\phantom{\rule{0.147em}{0ex}}(F)\propto \mathit{Rate}\mathit{of}\mathit{change}\mathit{of}\mathit{momentum}\\ \\ \mathit{Force}\phantom{\rule{0.147em}{0ex}}(F)\propto \phantom{\rule{0.147em}{0ex}}\frac{\mathit{Change}\mathit{in}\mathit{momentum}}{\mathit{time}}\\ \\ \mathit{Force}\phantom{\rule{0.147em}{0ex}}(F)\propto \phantom{\rule{0.147em}{0ex}}\frac{(m\times v)\; \u2013\; (m\times u)}{t}\\ \\ \mathit{To}\phantom{\rule{0.147em}{0ex}}\mathit{equate}\phantom{\rule{0.147em}{0ex}}\mathit{the}\phantom{\rule{0.147em}{0ex}}\mathit{terms},\phantom{\rule{0.147em}{0ex}}k(\mathit{proportionality}\mathit{constant})\phantom{\rule{0.147em}{0ex}}\mathit{is}\phantom{\rule{0.147em}{0ex}}\mathit{considered}\\ \\ \mathit{Force}\phantom{\rule{0.147em}{0ex}}(F)=\phantom{\rule{0.147em}{0ex}}k\times \left(\frac{(m\times v)\; \u2013\; (m\times u)}{t}\right)\\ \\ \mathit{Force}\phantom{\rule{0.147em}{0ex}}(F)=\phantom{\rule{0.147em}{0ex}}k\times \left(\frac{(m)\times (v-u)}{t}\right)\\ \\ \mathit{Here},k=1\mathit{in}\mathit{all}\mathit{systems}\mathit{of}\mathit{units}.\mathit{Hence},\\ \\ \mathit{Force}\phantom{\rule{0.147em}{0ex}}(F)=\phantom{\rule{0.147em}{0ex}}1\times \left(\frac{(m)\times (v-u)}{t}\right)\\ \\ \mathit{Force}\phantom{\rule{0.147em}{0ex}}(F)=\phantom{\rule{0.147em}{0ex}}\frac{(m)\times (v-u)}{t}\\ \\ \mathit{Since},\mathit{acceleration}=\frac{\mathit{change}\mathit{in}\mathit{velocity}}{\mathit{time}},\\ \\ a\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\frac{v-u}{t}.\mathit{Hence},\mathit{we}\mathit{have}\\ \\ F\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}m\phantom{\rule{0.147em}{0ex}}\times \phantom{\rule{0.147em}{0ex}}a\\ \\ \mathit{Force}=\mathit{mass}\times \mathit{acceleration}\end{array}$

No external force is needed to retain the motion of an object moving with uniform velocity. Whenever the net force acting on an object is not equal to zero, the velocity of the object will definitely change. Thus, a change in momentum occurs in the direction of the force. The change may happen either in magnitude or in direction, or in both.

Force is required to produce the acceleration of a body. In a uniform circular motion, even though the speed (magnitude of velocity) remains constant, the direction of the velocity changes at every point on the circular path. So, the acceleration is produced along the radius called centripetal acceleration. The force that produces this acceleration is called centripetal force, which you have learned in previous classes.