Useful constants Acceleration due to gravity=9.8m/s^2 Universal gravitational constant G=6.673times 10^-11Nm^2kg^-2 QUESTION ONE (30MARKS) a) Write down the units and dimensions of kinetic energy. (2mks) b) State Newtons second law of motion. (2mks) c) Abody A of mass 5kg moving with a velocity of 3mls collides head-on with another body B of mass 4kg moving in the opposite direction at 6mls. If after the collision the bodies move together, calculate the common velocity. (3mks) d) Two forces (500N,80^circ )and(600N,150^circ ) act at 9 point,determine the resultant force. (4mks) e) Distinguish between elastic and inelastic collision. (2mks) f) Explain why a body moving round a circular path at uniform tangential velocity is said to be accelerating. (2mks) g) Find the angular frequency of a car that moves at uniform velocity of 2mls in a circular path of 0.4m radius. Take pi =3.142 (4mks) Page 1 of 4 h) A car drives off the edge of a cliff that is 54m high. A police officer at the scene of accident notes that the point of impact is 150m from the base of the cliff. How fast was the car travelling when it went off the cliff? (5mks) i) Given that overrightarrow (n)=3i-2j-1k overrightarrow (0)=2j-4i+3k Find overrightarrow (n)cdot overrightarrow (0) (3mks) j) A stone of mass 0.6kg attached to a string of length 0.5m, is whirled in a horizontal circle at a constant speed. If the maximum tension in the string is 30N before it breaks. Calculate the maximum speed of the stone. (3mks) a) (i) What is moment of intertia (1mk) (ii) A fly wheel of radius 0.75m has a mass of 250kg and a torque of 30N acts on it. Calculate its moment of inertia and angular accelerated. (6mks) b) For a uniformly accelerated body derive the equation for the kinetic energy gained. (6mks) c) For the gravitational intensity g show that the period T is given by T=2pi sqrt (4g) (7mks) QUESTION THREE (20 MARKS) a) A metal rod of radius 4.0mm and length 90cm is held to a clamp.A force of magnitude 90,000N is applied perpendicularly to the end face at the other end. If the elongation is 2.0mm ,determine the Youngs modulus of elasticity of the rod.

QUESTION ONE (30 MARKS)<br /><br />a) Units and dimensions of kinetic energy:<br /> - Units: Joules (J)<br /> - Dimensions: [M][L]^2[T]^-2<br /><br />b) Newton's second law of motion:<br /> - Newton's second law of motion states that the force acting on an object is equal to the mass of the object multiplied by its acceleration. Mathematically, it is expressed as F = ma, where F is the force, m is the mass, and a is the acceleration.<br /><br />c) Common velocity after collision:<br /> - To find the common velocity after the collision, we can use the conservation of momentum. The total momentum before the collision is equal to the total momentum after the collision.<br /> - Initial momentum of body A: p_A = m_A * v_A = 5 kg * 3 m/s = 15 kg*m/s<br /> - Initial momentum of body B: p_B = m_B * v_B = 4 kg * (-6 m/s) = -24 kg*m/s<br /> - Total initial momentum: p_total_initial = p_A + p_B = 15 kg*m/s - 24 kg*m/s = -9 kg*m/s<br /> - After the collision, the bodies move together with a common velocity v_common. The total momentum after the collision is: p_total_final = (m_A + m_B) * v_common<br /> - Equating the total initial momentum to the total final momentum: -9 kg*m/s = (5 kg + 4 kg) * v_common<br /> - Solving for v_common: v_common = -9 kg*m/s / 9 kg = -1 m/s<br /><br />d) Resultant force:<br /> - To find the resultant force, we can use the method of vector addition. We need to break down each force into its horizontal and vertical components and then add them up.<br /> - Force 1: (500 N, 80°) - Horizontal component: 500 N * cos(80°) = 91.6 N, Vertical component: 500 N * sin(80°) = 475.9 N<br /> - Force 2: (600 N, 150°) - Horizontal component: 600 N * cos(150°) = -299.2 N, Vertical component: 600 N * sin(150°) = 455.4 N<br /> - Resultant force: (91.6 N - 299.2 N, 475.9 N + 455.4 N) = (-207.6 N, 931.3 N)<br /><br />e) Distinguish between elastic and inelastic collision:<br /> - Elastic collision: In an elastic collision, both momentum and kinetic energy are conserved. The total kinetic energy before the collision is equal to the total kinetic energy after the collision.<br /> - Inelastic collision: In an inelastic collision, only momentum is conserved, but kinetic energy is not conserved. Some of the kinetic energy is converted into other forms of energy, such as heat or sound.<br /><br />f) Explanation of acceleration in circular motion:<br /> - A body moving round a circular path at uniform tangential velocity is said to be accelerating because acceleration is a vector quantity, and it includes both changes in speed and direction. Even if the speed remains constant, the direction of the velocity is constantly changing, resulting in a centripetal acceleration directed towards the center of the circle.<br /><br />g) Angular frequency:<br /> - Angular frequency (ω) is given by the formula ω = v / r, where v is the tangential velocity and r is the radius of the circular path.<br /> - Given: v = 2 m/s, r = 0.4 m<br /> - Angular frequency: ω = 2 m/s / 0.4 m = 5 rad/s<br /><br />h) Speed of the car:<br /> - We can use the kinematic equations to find the speed of the car. Specifically, we can use the equation for projectile motion: h = (1/2) * g * t^2, where h is the height, g is the acceleration due to gravity, and t is the time.<br /> - Given: h = 54 m, horizontal distance = 150 m<br /> - Time of flight: t = sqrt(2 * h / g) = sqrt(2 * 54 m / 9.8 m/s^2) = 3.36 s<br /> - Horizontal velocity: v_x = horizontal distance / t = 150 m / 3.36 s = 44.64 m/s<br /><br />i) Dot product of vectors:<br /> - Given: $\overrightarrow{n} = 3i - 2j - 1k$, $\overrightarrow{0} = 2j - 4i + 3k$<br /> - Dot product: $\overrightarrow{n} \cdot \overrightarrow{0} = (3 * -4) + (-2 * 2) + (-1 * 3) = -12 - 4 - 3 = -19$<br /><br />j) Maximum speed of