A monkey of mass \(20 text{ kg}\) is holding a vertical rope. The rope will not break when a mass of \(25 text{ kg}\) is suspended from it but will break if the mass exceeds \(25 text{ kg}\). What is the maximum acceleration with which the monkey can climb up along the rope? (\(g = 10 text{ m/s}^2\)) (2003)
Monkey mass \(m = 20 text{ kg}\). Max tension \(T_{max} = 25 text{ kg} times g = 250 text{ N}\).When climbing up, \(T - mg = ma\).\(250 - 20(10) = 20a implies 250 - 200 = 20a\).So, \(50 = 20a implies a = 2.5 text{ m/s}^2\).
A lift of mass \(1000 text{ kg}\) which is moving with acceleration of \(1 text{ m s}^{-2}\) in upward direction, then the tension developed in string which is connected to lift is: (2002)
A man is slipping on a frictionless inclined plane and a bag falls down from the same height. Then the speed of both is related as: (2000)
For the bag, \(v_B = sqrt{2gh}\).For the man on a frictionless incline, acceleration \(a_m = g sintheta\). If he slides a distance \(L\), then \(h = L sintheta\).Speed \(v_m = sqrt{2 a_m L} = sqrt{2 (g sintheta) (h/sintheta)} = sqrt{2gh}\).Thus, \(v_B = v_m\).
A small ball is suspended from a thread. It is lifted up with an acceleration \(4.9 text{ m s}^{-2}\) and lowered with an acceleration \(4.9 text{ m s}^{-2}\) then the ratio of tensions in the thread in both cases will be: (1998)
A monkey is descending from branch of a tree with constant acceleration. If the breaking strength is \(75%\) of the weight of the monkey, the minimum acceleration with which monkey can slide down without branch is: (1993)
Let monkey's mass be \(m\). Weight \(W = mg\). Max tension \(T_{max} = 0.75 mg\).When sliding down, \(mg - T = ma\). For minimum acceleration, \(T = T_{max}\).So, \(mg - 0.75mg = ma implies 0.25mg = ma\).Thus, \(a = 0.25g = g/4\).
A shell of mass \(m\) is at rest initially. It explodes into three fragments having mass in the ratio \(2 : 2 : 1\). If the fragments having equal mass fly off along mutually perpendicular directions with speed \(v\), the speed of the third (lighter) fragment is: (2022)
A person of mass \(60 text{ kg}\) is inside a lift of mass \(940 text{ kg}\) and presses the button on control panel. The lift starts moving upwards with an acceleration \(1.0 text{ m/s}^2\). If \(g = 10 text{ m/s}^2\), the tension in the supporting cable is (2011 Pre)
A ball of mass \(0.15 text{ kg}\) is dropped from a height \(10 text{ m}\), strikes the ground and rebounds to the same height. The magnitude of impulse imparted to the ball is (\(g = 10 text{ m/s}^2\)) nearly: (2021)
Speed before impact \(v = sqrt{2gh} = sqrt{2 times 10 times 10} = 10sqrt{2} text{ m/s}\).Since it rebounds to the same height, speed after impact is also \(v = 10sqrt{2} text{ m/s}\).Impulse \(J = Delta P = m(v_{final} - v_{initial})\). Considering upward as positive, \(J = m(v - (-v)) = 2mv\).\(J = 2 times 0.15 times 10sqrt{2} = 3sqrt{2} approx 4.24 text{ Ns}\).
A stone is dropped from a height \( h \). It hits the ground with a certain momentum \( P \). If the same stone is dropped from a height \( 100\% \) more than the previous height, the momentum when it hits the ground will change by: (2012 Mains)
Momentum \( P = msqrt{2gh} \). If height increases by \( 100\% \), new height \( h' = 2h \). New momentum \( P' = msqrt{2g(2h)} = sqrt{2}P \). Percentage change \( \frac{P' - P}{P} \times 100\% = (sqrt{2} - 1)\times 100\% \approx 41.4\% \).
A body of mass \( m \) hits normally a rigid wall with velocity \( v \) and bounces back with the same velocity. The impulse experienced by the body is: (2011 Pre)
Impulse is the change in momentum. Initial momentum \( P_i = mv \). Final momentum \( P_f = -mv \) (opposite direction). Impulse \( J = P_f - P_i = -mv - mv = -2mv \). Magnitude is \( 2mv \).