Question 91:
A man fires a bullet of mass \( 200 \text{ gm} \) at a speed of \( 5 \text{ m/s} \). The gun is of one \( text{kg} \) mass. By what velocity the gun rebounds backward? (1996)
By conservation of momentum, initial momentum = final momentum. \( 0 = m_b v_b + m_g v_g \). \( 0 = (0.2 \text{ kg})(5 \text{ m/s}) + (1 \text{ kg}) v_g \). \( 0 = 1 + v_g \). \( v_g = -1 \text{ m/s} \). The gun rebounds with \( 1 \text{ m/s} \).
Question 92:
A satellite in force free space sweeps stationary interplanetary dust at a rate of \( dM/dt = \alpha v \), where \( M \) is mass and \( v \) is the speed of satellite and \( alpha \) is a constant. The acceleration of satellite is: (1995)
The force on the satellite due to dust accretion is \( F = v_{text{rel}} \frac{dM}{dt} \). Here, the dust is stationary, so \( v_{text{rel}} = -v \). So, \( F = (-v) \frac{dM}{dt} \). From Newton's second law, \( F = M \frac{dv}{dt} \). Thus, \( M \frac{dv}{dt} = -v \frac{dM}{dt} \). Substituting \( \frac{dM}{dt} = \alpha v \), we get \( M \frac{dv}{dt} = -v (\alpha v) = -\alpha v^2 \). Therefore, \( \frac{dv}{dt} = -\frac{\alpha v^2}{M} \).
Question 93:
In a rocket, fuel burns at the rate of \( 1 \text{ kg/s} \). This fuel is ejected from the rocket with a velocity of \( 60 \text{ km/s} \). This exerts a force on the rocket equal to: (1994)
Thrust force \( F = v_e \frac{dM}{dt} \). Given exhaust velocity \( v_e = 60 \text{ km/s} = 60000 \text{ m/s} \). Given rate of fuel consumption \( \frac{dM}{dt} = 1 \text{ kg/s} \). \( F = 60000 \text{ m/s} \times 1 \text{ kg/s} = 60000 \text{ N} \).
Question 94:
A body of mass \( 5 \text{ kg} \) explodes at rest into three fragments with masses in the ratio \( 1 : 1 : 3 \). The fragments with equal masses fly in mutually perpendicular directions with speeds of \( 21 \text{ m/s} \). The velocity of heaviest fragment in \( text{m/s} \) will be: (1989)
Masses \( m_1 = 1 \text{ kg}, m_2 = 1 \text{ kg}, m_3 = 3 \text{ kg} \). Momentum of first two parts: \( P_1 = 1 \text{ kg} \times 21 \text{ m/s} = 21 \text{ Ns} \), \( P_2 = 1 \text{ kg} \times 21 \text{ m/s} = 21 \text{ Ns} \). Since they are perpendicular, their resultant momentum \( P_{12} = sqrt{21^2 + 21^2} = 21sqrt{2} \text{ Ns} \). By conservation of momentum, \( P_3 = P_{12} = 21sqrt{2} \text{ Ns} \). Velocity of heaviest part \( v_3 = P_3 / m_3 = (21sqrt{2}) / 3 = 7sqrt{2} \text{ m/s} \).
Question 95:
A block of mass (10text{ kg}) is in contact against the inner wall of a hollow cylindrical drum of radius (1text{ m}). The coefficient of friction between the block and the inner wall of the cylinder is 0.1. The minimum angular velocity needed for the cylinder to keep the block stationary when the cylinder is vertical and rotating about its axis, will be : ((g = 10text{ m/s}^2)) (2019)
The normal force (N) provides centripetal force, so (N = momega^2 R). For the block not to fall, friction (f = mu N) must balance its weight (mg). Thus, (mu momega^2 R = mg). This simplifies to (omega^2 = frac{g}{mu R}). Substituting values: (omega = sqrt{frac{10}{0.1 times 1}} = sqrt{100} = 10text{ rad/s}).
Question 96:
One end of string of length (l) is connected to a particle of mass ‘m’ and the other end is connected to a small peg on a smooth horizontal table. If the particle moves in circle with speed ‘v’, the net force on the particle (directed towards center) will be: (T represents the tension in the string) (2017-Delhi)
The particle is undergoing uniform circular motion. The net force directed towards the center is the centripetal force. In this case, the tension (T) in the string is the sole force providing the centripetal force. Therefore, the net force on the particle, directed towards the center, is simply (T).
Question 97:
A car is negotiating a curved road of radius R. The road is banked at an angle (theta). The coefficient of friction between the tyres of the car and the road is (mu_s). The maximum safe velocity on this road is: (2016 – I)
The maximum safe velocity on a banked road with friction is given by the formula (v = sqrt{gR frac{mu_s + tantheta}{1 - mu_s tantheta}}\) by resolving normal and friction forces into horizontal and vertical components. The horizontal components contribute to centripetal force while vertical components balance weight.
Question 98:
Two stones of masses (m) and (2m) are whirled in horizontal circles, the heavier one in a radius (r/2) and the lighter one in radius (r). The tangential speed of lighter stone is (n) times that of the value of heavier stone when they experience same centripetal forces. The value of (n) is: (2015 Re)
Centripetal force (F_c = frac{mv^2}{r}). Given (F_{c1} = F_{c2}). So (frac{(2m)v_1^2}{(r/2)} = frac{m v_2^2}{r}). This simplifies to (frac{4mv_1^2}{r} = frac{mv_2^2}{r}), which gives (v_2^2 = 4v_1^2), so (v_2 = 2v_1). Therefore, (n=2).
Question 99:
A car of mass m is moving on a level circular track of radius R. If (mu_s) represents the static friction between the road and tyres of the car, the maximum speed of the car in circular motion is given by: (2012 Mains)
On a level circular track, the maximum static friction provides the necessary centripetal force. So, (frac{mv_{max}^2}{R} = mu_s mg). Canceling (m), we get (v_{max}^2 = mu_s gR). Therefore, (v_{max} = sqrt{mu_s gR}).
Question 100:
Starting from rest, a body slides down a (45^circ) inclined plane in twice the time it takes to slide down the same distance in the absence of friction. The coefficient of friction between the body and the inclined plane is: (1988)
Acceleration without friction is (a_1 = gsintheta). With friction, (a_2 = g(sintheta - mucostheta)). Since (s = frac{1}{2}at^2), (a_1 t_1^2 = a_2 t_2^2). Given (t_2 = 2t_1), so (a_1 t_1^2 = a_2 (2t_1)^2 = 4 a_2 t_1^2). Thus, (a_1 = 4a_2). (gsintheta = 4 g(sintheta - mucostheta)). (sintheta = 4sintheta - 4mucostheta). So, (4mucostheta = 3sintheta), giving (mu = frac{3}{4}tantheta). For (theta=45^circ), (mu = frac{3}{4} times 1 = 0.75).