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Laws of Motion — Physics STD 11 Science — Question

Gujarat 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 $m_1$ and $m_2$ 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 $m_1$ moves downward with an acceleration a.
The resultant downward force while considering the heavier mass m, is given by
$m_1g - T = m_1a ...(i)$
The resultant upward force while considering the lighter mass $m_2$ is given by
$T - m_2g = m_2a ...(ii)$
Adding (i) and (ii), we have
$(m_1 - m_2)g = (m_1 + m_2)a$
$\Rightarrow\ \text{a}=\Big(\frac{\text{m}_1-\text{m}_2}{\text{m}_1+\text{m}_2}\Big)\text{g}\dots(\text{iii})$
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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