#$&* course Phy 121 Test Problems: Problem Number 1 The velocity of an object of constant mass 6 Kg is observed to change from 3 m/s to -129 m/s in 6 seconds. Use the Impulse-Momentum Theorem to determine the average force exerted on the object. Verify your results using your knowledge of uniformly accelerated motion. Impulse- Momentum `dv = -129m/s - 3m/s = F*`dt = m *`dv F *`6s =6kg * -132m/s F= (-132m/s *6kg )/6s = -132 kg m/s^2 or -132N Using Uniform Accelerated Motion a = `dv/`dt = (-132m/s) / 6 s a= -22m/s^2 fAve = m*a fAVe =6kg * -22m/s * = -132kg m/s^2 = -132N Verified. . . . . . . . . . Problem Number 2 Assuming that 10360 Joules of kinetic energy are dissipated in the process, how much kinetic energy increase should be expected as a result of a 70 Newton force being exerted over a distance of 200 meters? `dKE = 70N * 200 m = 14000 N m = 14000 J 14000J - 10360 J = 3640 J . . . . . . . . . . Problem Number 3 An object is moving at 6 meters per second. If its mass is 3 kilograms, what is its kinetic energy? If its speed doubled, what would be its kinetic energy? How many times the original kinetic energy would it have after its speed doubled? KE = .5m *v^2 KE = .5 * 3kg * (6m/s) ^2 = 54 kg *m/s^2 = 54 J Double speed KE = .5m *v^2 KE = .5 * 3kg * (12m/s) ^2 = 216 kg *m/s^2 = 216 J After doubling the speed, it has 216J /54J or 4 times the original kinectic energy. . . . . . . . . . . Problem Number 4 An object moving to the right at 3 m/s collides with an object moving at -9 m/s (i.e., to the left). After the collision the first object is observed to have velocity -3 m/s (this negative velocity is toward the left). The mass of the first object is 4 kg and the mass of the second is 6 kg. The duration of the collision is .076 seconds. What was the average force exerted by the second object on the first? What was the average force exerted by the first object on the second? What will be the velocity of the second object after the collision? How do the kinetic energy totals before collision compare with those after collision? `dp1 = m1 *`dv1 `dp1 =4kg * (-3m/s - 3m/s) `dp1 = 4kg * -6m/s = `dp1 = -24 kg m/s Fave 12 = `dp1/`dt = (-24kg m/s) / .076s = -316 kg m/s^2 approx. Fave 21 is equal and opposite Fave 12 so it is 316kg m/s^2 `dp2 = Fave 21 *`dt `dp2 = 316kg m/s *.076 s = 24 kg m/s `dp2= `dv2 * m2 `dv2= dp2/m2 `dv2 =( 24kg m/s) / 6kg = 4m/s vf = v0 + dv2 = -9m/s + 4m/s = -5m/s Velocity of second object will be -5m/s after the collision. KEbefore = .5 * 4kg (3m/s)^2 + .5 * 6kg (-9m/s)^2 KE before = 18kg m^2/s^2 + 243 kg m^2/s^2 = 261 Joules KE after = .5 * 4kg (-3m/s)^2 + .5 * 6kg (-5m/s)^2 KE after = 18kg m^2/s^2 + 75 kg m^2/s^2 = 93 Joules KEafter is less than KEbefore . . . . . . . . . . Problem Number 5 How much force, exerted parallel to the direction of motion, is necessary when pushing an object 8 meters in order to use up 152 Joules of energy from the pushing source? KE = F * `ds = F * 8m F*8m = 152 N meters F = 152 N meters/8m = 19 N . . . . . . . . . . Problem Number 6 An object of mass 15 kg experiences a variable force F(t) (here F(t) indicates function notation, not multiplication of F by t; F(t) is the force at clock time t) for .1 seconds. If the average force over this time is 795 Newtons, Use Newton's Second Law and your knowledge of uniformly accelerated motion to find the change in the object's velocity. Use the Impulse-Momentum Theorem to obtain the same result. F Ave = m *a a = F Ave/m a = (795 kg m/s^2 )/15 kg = 53m/s^2 `dv = a * `dt `dv = 53m/s^2 * .1s = 5.3m/s Using the Impulse-Momentum Theorem F * `dt = m *`dv 795 kg m/s^2 * .1 s = 15kg * `dv `dv = (795 kg m/s^2 * .1 s)/15kg `dv = 5.3 m/s Verified. . . . . . . . . . . Problem Number 7 What vector of magnitude 9.9 must be added to the velocity vector A = < -4.66 m/s, 4.15 m/s> in order to obtain a vertical vector R? Answer by giving the magnitude and angle of the vector to be added. (Note on notation: