Physics 1A, Week 2 Quiz Solutions

Transcription

1 Vector _ A points north and vector _ B points east. If _ C = _ B _ A, then vector _C points: a. north of east. b. south of east. c. north of west. d. south of west. Find the resultant of the following two vectors: i) 50 units due east and ii) 100 units 30 north of west. a. 100 units 30 north of west b. 6 units 15 north of west c. 87 units 60 north of west d. 6 units 54 north of west Solution: To solve this problem, we must break the second vector into NSEW components. The north component is 100*sin(30)=50 and the west component is 100*cos(30)=87. The resultant vector is 37 units to the west and 50 units to the north. The length of this vector is 6 units. The angle of this vector from the west direction is arctan(50/37) = 54. A baseball thrown from the outfield is released from shoulder height (1.5 m off the ground) at an initial velocity of 9.4 m/s at an initial angle of 30.0 with respect to the horizontal. What is the maximum vertical displacement that the ball reaches during its trajectory? a m b m c..1 m d m Solution: First we find the initial y-velocity, which is 9.4*sin(30) = 14.7 m/s upwards. Acceleration due to gravity is 9.8 m/s^ downward. At the top of the trajectory, the ball stops, so there vy=0. Plugging this into our kinematics equation v y -v 0y = a y (Δy) and solving for Δy, we find Δy = -(14.7m/s) /[*(-9.80 m/s )] = 11.0 m. Δy is the vertical displacement. A helicopter is traveling at 40 m/s at a constant altitude of 100 m over a level field. If a wheel falls off the helicopter, with what speed will it hit the ground? (g = 9.8 m/s and air resistance negligible) a. 40 m/s b. 50 m/s c. 60 m/s d. 70 m/s

2 Solution: Looking at motion in the y-direction first, we see that the initial y-velocity is 0 m/s and the its y-acceleration is the acceleration due to gravity (directed downward). Δy is -100 m since it falls from the initial height of off the ground to the ground. Therefore its y-velocity as it hits the ground can be found using v y -v 0y = a y (Δy). v y -(0 m/s) = (-9.8 m/s)(-100 m). v y = -44. m/s. Since the ball is heading towards the ground (downwards in our coordinate system), we choose the negative root. The total velocity takes into account both the x and y-velocities. To find the x-velocity, we note that there is no acceleration in the x-direction. Therefore the x-velocity stays constant at 40 m/s. The total velocity, using the Pythagorean theorem for the x and y velocities is 60 m/s. A rifle is aimed horizontally toward the center of a target 100 m away. If the bullet strikes 10 cm below the center, what was the velocity of the bullet? (Ignore air friction.) a. 300 m/s b. 333 m/s c. 500 m/s d. 700 m/s Solution: First we find final y-velocity by analyzing motion in the y-direction. The initial velocity in the y-direction is 0 since the rifle is aimed horizontally, the acceleration in the y- direction is 9.8 m/s towards the earth (which we choose to call downward here), and the y- displacement (Δy) is -10 cm (since the final height is beneath the initial height). Substituting these values into our kinematics equation v y -v 0y = a y (Δy), we find v y = (-9.8 m/s ) (-0.10 m) v y = -1.4 m/s The time it takes to hit the target can be found by either of our other kinematics equations: v y = v 0y + a y t t = ( v y - v 0y) /a y = (-1.4 m/s)/(-9.8 m/s ) = 0.14 s We now use this time to find the x-velocity, knowing that the x-acceleration is 0: Δx = v 0x t +(1/)a x t v 0x =(Δx)/t = 100 m / 0.14 s = 700 m/s

3 Since there is no x-acceleration, the initial x-velocity is the same as the final x-velocity. To find the total velocity, we should use the Pythagorean theorem and substitute in the x and y velocities. v = (700 + (-1.4) ) 700 m/s. The acceleration due to gravity on the Moon s surface is one-sixth that on Earth. What net force would be required to accelerate a 0-kg object along the moon's surface at 6.0 m/s? a. 1.3 N b. 0 N c. 33 N d. 10 N Solution: Net force is always equal to total mass times net acceleration. Here we are given that the total acceleration should be 6.0 m/s and the mass is 0 kg. Multiplying the two, we find that the total force is 10 N. An automobile of mass 000 kg moving at 30 m/s is braked suddenly with a constant braking force of N. How far does the car travel before stopping a. 45 m b. 90 m c. 135 m d. 180 m Solution: First we find the car's acceleration by dividing the total force on it by its mass. Therefore a = N/ 000 kg = 5 m/s. Because the braking force is constant, the acceleration is also constant. This is important because we can only use the kinematic equations when acceleration is constant. To find out how far the car traveled, we use our kinematics equation relating x-acceleration, displacement, and velocities. The final x-velocity is 0 since it stops, the initial x-velocity is 30 m/s, and the x-acceleration is -5 m/s since the car is slowing down and we chose to make the initial x-velocity positive. Plugging this in, v x -v 0x = a x (Δx) (Δx) = -v 0x /( a x ) = - (30 m/s) /( (-5 m/s ) ) = 90 m Two blocks of masses 0 kg and 8 kg are connected together by a light string and rest on a frictionless level surface. Attached to the 8-kg mass is another light string, which a person

4 uses to pull both blocks horizontally. If the two-block system accelerates at 0.5 m/s what is the tension in the connecting string between the blocks? a. 14 N b. 6 N c. 10 N d. 4.0 N Solution: The 0 kg mass has only one force in the horizontal direction: the tension. Therefore, the net force on this mass, which is always equal to its mass * acceleration, is F = (0 kg)*(0.5 m/s) = 10 N. A 0-kg traffic light hangs midway on a cable between two poles 40 meters apart. If the sag in the cable is 0.40 meters, what is the tension in each side of the cable? a N d. 980 N b N c N d. 980 N Solution: T_1 T_ We choose x to be parallel to the bottom of the page and pointing to the right and y to be positive upwards. F_g The light is in equilibrium, so there are no net forces on it. and y directions: We add up the forces in both the x F_Net,x = T_x - T_1x = m*a_x=0. The x-components of the two tensions are equal. F_Net,y = T_1y + T_y - F_g = m*a_y=0. Assuming the problem is symmetric, T_1y = T_y. T_1y = (1/) F_g = (1/)(0 kg)(9.8 m/s^) Using the sag and half the distance between the poles to make a triangle, 0.4 m 0 m

5 If theta is the angle between the bottom and the hypotenuse, tan theta = (0.4/0), so theta = T_1 = T_1y/sin theta = N. A block of mass 5.00 kg rests on a horizontal surface where the coefficient of kinetic friction between the two is A string attached to the block is pulled horizontally, resulting in a.00-m/s acceleration by the block. Find the tension in the string. (g = 9.80 m/s ) a N b N c N d N Solution: F_N (normal force) F_friction Tension F_g (gravity) We choose the x-axis to be on the horizontal and positive to the right. Therefore, a_x = m/s^ We choose the y-axis to be on the vertical and positive upward. F_net,x = T F_friction = m*a_x. = (5 kg)*( m/s^) = 10 N Solving for T, t = F_Net,x + F_friction. Now we need to find F_friction. F_net,y = F_N F_g = m*a_y The block doesn t move up or down off the table, so a_y = 0. This means F_N = F_g. F_friction = µ*f_n = µ*f_g = µmg = (.)(5)(9.8) = 9.8 N. Therefore, T = 10 N N = 19.8 N