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A 7 kg object is subjected to two forces (in newton) $ \vec F _1 = 20 \hat i + 30 \hat j $ and $ \vec F_2 = 8 \hat i -5 \hat j $. The magnitutde of resulting acceleration in $ ms^{-2} $ will be
$ \vec F = \vec F_1 + \vec F_2 $ $ \therefore a = F /m $
A car travelling at a speed of 30 km/h is brought to a halt in 8 metres by applying brakes. If the same car is travelling at 60 km/h it can be brought to a halt with the same breaking power in
Here distance $ d \alpha \nu^2 $ as $ \nu $ is doubled, d becomes 4 times.
A given object takes n times more time to slide down $ 45 ^\circ $ rough inclined plane as it takes to slide down a perfactly smooth $ 45 ^\circ $ incline. The coefficient of kinetic friction between the object and the incline is
$ S = ut + { 1 \over 2 } at^2 $ $as u = 0 ,t = \sqrt { 2S \over a } $ for smooth plane $a = gsin \theta $ For rough plane$ a^1= g (sin \theta -\mu cos \theta)$ $ \therefore t^1 = \sqrt { 2S \over g ( sin \theta - \mu cos \theta ) } $ $ \therefore t^1 = nt = n \sqrt { 2S \over g sin \theta } $ $ \therefore n^2 g (sin \theta - \mu cos \theta) = g sin \theta $ $ when \theta = 45 sin \theta = cos \theta = { 1 \over \sqrt 2 } $ $ \therefore = 1 - { 1 \over n^2 } $
Two bodies of equal masses revolve in circular orbits of radii R1 and R2 with the same period Their centripetal forces are in the ratio
$ F + { mv^2 \over R } = { m(R \omega )^2 \over R } = mR \left({ 2 \pi \over T } \right) ^2 = { 4 \pi ^2 m R \over T^2 } $ As m and T are same for two bodies $ F \alpha R $
A 0.5 kg ball moving with a speed of $12 ms^{-1} $ strikes a hard wall at an angle of 300 with the wall. It is reflected with the same speed and at the same angle. If the ball is in contact with the wall for 0.025 S the average force acting on the wall is
As impulse = change in linear momentum $ Ft = 2 mv sin 30 \Rightarrow F = { 2 m \vartheta sin 30 \over t } $
A shell of mass 200g is ejected from a gun of mass 4 kg by an explosion that generates 1.05 KJ of energy. The initial velocity of the shell is
$ m_1 v_1 + m_2 v_2 = 0 $ $ v_2 = { m_1 \over m_2 } v_1 = - { v_1 \over 20 } $ Now $ E = { 1 \over 2 } m_1 v_1^2 + {1 \over 2 } m_2 v_2^2 $ By solving weight $ \nu_1 = 100 ms^{-1} $
A gramophone record is revolving with an anguler velocity $ \omega $. A coin is placed at a distance r from the centre of the record. The coefficient of static friction is . The coin will revolve with the record if
The coin will revolve with the record if centrifugal force $ mrw^2 \lt f $ $ \therefore mr \omega^2 \lt R \lt mg $ $ r \lt { g \over \omega ^2 } $
A stone of mass 2 k g is tied to a string of length 0.5 m It the breaking tension of the string is 900N, then the maximum angular velocity the stone can have in uniform circular motion is
$ F = mr \omega^2 \Rightarrow \omega = \sqrt { F /mr } = 30 rad/s $
A sparrow flying in air sits on a stretched telegraph wire. If the weight of the sparrow is W ,which of the following is true about the tension T produced in the wire?
When the sparrow sits on the wire, it causes a deflection in the wire, creating a vertical component of tension. The tension in the wire must counteract not only the weight of the sparrow but also the additional vertical components due to the deflection. Therefore, the tension T in the wire is greater than the weight W of the sparrow: $T > W$.
A body of mass 0.05 kg is falling with acceleration $ 9.4 ms^{–2} $ . The force exerted by air opposite to motion is …………. N $(g=9.8 ms^{–2})$
Fair = m ( g - a)
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