Chapter 5
Applied Mathematics · 15 exercises
Problem 1
A particle of mass \(5 \mathrm{~kg}\) is pulled along a smooth horizontal surface by a horizontal string. The acceleration of the particle is \(10 \mathrm{~ms}^{-2}\). The tension in the string is: (a) \(2 \mathrm{~N}\) (b) \(50 \mathrm{~N}\) (c) \(5 \mathrm{~N}\) (d) \(15 \mathrm{~N}\) (e) \(10 \mathrm{~N}\).
5 step solution
Problem 2
A particle of mass \(3 \mathrm{~kg}\) slides down a smooth plane inclined at arcsin \(\frac{1}{3}\) to the horizontal. The acceleration of the particle is: (a) \(\frac{R}{3} \mathrm{~ms}^{-2}\) (b) \(g \mathrm{~ms}^{-2}\) (c) \(1 \mathrm{~ms}^{-2}\) (d) \(3 g \mathrm{~ms}^{-2}\) (e) 0 .
5 step solution
Problem 3
A block of mass \(10 \mathrm{~kg}\) rests on the floor of a lift which is accelerating upwards at \(4 \mathrm{~ms}^{-2}\). The reaction of the floor of the lift on the block is: (a) \(104 \mathrm{~N}\) (b) \(96 \mathrm{~N}\) (c) \(60 \mathrm{~N}\) (d) \(30 \mathrm{~N}\) (e) \(140 \mathrm{~N}\).
4 step solution
Problem 5
Two particles of mass \(3 \mathrm{~kg}\) and \(5 \mathrm{~kg}\) are connected by a light inextensible string passing over a smooth pulley which is fixed to the ceiling of a lift. Find the tension in the string when the system is moving freely and the lift has a downward acceleration \(g \mathrm{~ms}^{-2}\).
5 step solution
Problem 7
A particle is moving with uniform velocity. (a) The forces acting on the particle are in equilibrium. (b) The particle has zero acceleration. (c) There is a resultant force acting on the body in the direction of the velocity.
4 step solution
Problem 8
A particle is moving horizontally with constant acceleration. (a) The sum of the horizontal components of the forces acting on the particle is not zero. (b) The sum of the vertical components of the forces acting is not zero. (c) The forces acting on the particle are not in equilibrium.
4 step solution
Problem 9
The diagram shows a smooth, weightless pulley suspended by a light support AB. A light, inextensible string passes over the pulley and carries masses of \(5 \mathrm{~kg}\) and \(3 \mathrm{~kg}\) at its ends. The masses are released from rest. Find the acceleration of each mass in \(\mathrm{ms}^{-2}\) and the tension in the string in newtons. The support \(\mathrm{AB}\) is now made to descend with an acceleration of \(3 \mathrm{~ms}^{-2}\). Find the magnitude and direction of the acceleration in space of each of the masses and the new tension in the string. (Cambridge)
8 step solution
Problem 10
Two particles A and B of masses 3 and \(4 \mathrm{~kg}\) are connected by a light inelastic string passing over a smooth fixed pulley. (a) The acceleration of A is \(\frac{g}{7} \mathrm{~ms}^{-2}\) upwards. (b) The tension in the string is \(\frac{24}{7} g \mathrm{~N}\). (c) The acceleration of \(\mathrm{B}\) is \(-\frac{g}{7} \mathrm{~ms}^{-2}\) upwards.
6 step solution
Problem 12
A light inextensible string passes over a smooth fixed pulley and has a particle of mass \(5 m\) attached to one end and a second smooth pulley of mass \(m\) attached to the other end. Another light inextensible string passes over the second pulley and carries a mass \(3 \mathrm{~m}\) at one end and a mass \(m\) at the other end. If the system moves freely under gravity, find the acceleration of the heaviest particle and the tension in each string. (U of \(\mathrm{L})\)
5 step solution
Problem 14
(a) A particle is moving with a constant velocity. (b) The forces acting on a particle are in equilibrium.
4 step solution
Problem 15
Two wooden discs \(\mathrm{X}\) and \(\mathrm{Y}\) of thickness \(2 a\) and \(4 a\) respectively are fixed at a small distance apart with their plane faces vertical and parallel. A small bullet of mass \(m\) is fired horizontally into \(X\) with initial speed \(u\) at right angles to the plane faces. It emerges from \(X\) with speed \(v\) and then enters \(Y\) into which it penetrates a distance \(a\) before being brought to rest. If the motion of a bullet through \(\mathrm{X}\) and \(\mathrm{Y}\) is opposed by constant forces \(R_{1}, R_{2}\) respectively, find expressions for \(R_{1}\) and \(R_{2}\) in terms of \(u, v, a\) and \(m\). A second bullet of mass \(m\) is now fired horizontally into \(Y\) with initial speed \(u\) at right angles to the plane faces in a direction towards \(\mathrm{X}\). Show that this bullet will emerge from \(\mathrm{Y}\) if \(v<\frac{1}{2} u\). If \(v=\\{u\) and the second bullet enters \(X\) after emerging from \(Y\), find the distance which it penetrates into \(\mathrm{X}\) before being brought to rest. (The effect of gravity may be ignored). (Cambridge)
5 step solution
Problem 16
Two points \(\mathrm{A}\) and \(\mathrm{B}\) on a rough horizontal table are at a distance \(a\) apart. \(\mathrm{A}\) particle is projected along the table from A towards \(\mathrm{B}\) with speed \(u\), and simultaneously another particle is projected from B towards A with speed \(3 u\). The coefficient of friction between each particle and the table is \(\mu\). By considering the distance travelled by each particle before coming to rest, show that the particles collide if \(u^{2} \geqslant \frac{1}{3} \mu\) ag. If \(u^{2}=\frac{4}{7} \mu a g\), show that the collision occurs after a time \([a /(7 \mu g)]^{\frac{1}{2}}\) and at a distance \(\frac{3}{14} a\) from \(A\). (Cambridge)
7 step solution
Problem 16
A ring is free to slide down a rough straight wire. Find the acceleration of the ring. (a) The coefficient of friction between the wire and the ring is \(\mu\). (b) The wire is inclined at an angle \(\theta\) to the horizontal. (c) The mass of the ring is \(m\).
5 step solution
Problem 20
A car is brought to rest by the action of its brakes which are assumed to exert a constant force on the car. Find the distance moved by the car before coming to rest. (a) The mass of the car is \(750 \mathrm{~kg}\). (b) The initial velocity of the car is \(40 \mathrm{~ms}^{-1}\). (c) The time taken is 5 seconds.
4 step solution
Problem 22
The S.I. unit of force \((N)\) is that force which will give a body of mass \(1 \mathrm{~kg}\) an acceleration of \(1 \mathrm{~ms}^{-2}\).
4 step solution