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Laws of Motion — Physics STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 SciencePhysicsLaws of Motion3 Marks
Question
  1. Explain the term impulse. Show that impulse of a variable force is equal to the area enclosed by the force-time curve.
  2. Two masses 8kg and 12kg are connected at the two ends of a light inextensible string that passes over a frictionless pulley. Find the acceleration of the masses and tension in the string, when the masses are released.

✓

Answer

  1. The product of the force and the time interval on which it acts or change in momentum is called impulse.

$\vec{\text{I}}-\vec{\text{F}}_\text{av}\times\text{t}=\text{dp}=\text{F.dt}=$ Area of shaded portion

Net impulse $=\int\limits_{\text{t}_1}^{\text{t}_2}\vec{\text{F}}\times\text{dt}=$ area under graph ABC and time axis.

  1. Consider two masses m1 and m2 are connected to the ends of an inextensible string passing over a smooth frictionless pulley.

Let T be the tension in the string which is uniform throughout.

The heavier mass m1 moves downward with an acceleration a.

The resultant downward force while considering the heavier mass m, is given by

m1g - T = m1a ...(i)

The resultant upward force while considering the lighter mass m2 is given by

T - m2g = m2a ...(ii)

Adding (i) and (ii), we have

(m1 - m2)g = (m1 + m2)a

$\Rightarrow\ \text{a}=\Big(\frac{\text{m}_1-\text{m}_2}{\text{m}_1+\text{m}_2}\Big)\text{g}\dots(\text{iii})$

Putting the value of a in (i) and simplifying we get

$\text{T}=\frac{2\text{m}_1\text{m}_2}{\text{m}_1+\text{m}_2}\text{g}\dots(\text{iv})$

Substituting the given values, we have

$\Rightarrow\ \text{a}=\Big(\frac{12-8}{12+8}\Big)\times9.8=1.96\text{ms}^{-2}$

$\text{T}=\frac{2\times12\times8\times9.8}{12+8}=\frac{1881.6}{20}$

$=64.08\text{N}$

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