BROCK UNIVERSITY SOLUTIONS. 1. [1 point] A car is driving at a constant speed on a circular road. The force on a passenger in the car is

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1 BROCK UNIVERSITY Test 2: October 2014 Number of pages: 4 + formula sheet Course: PHYS 1P21/1P91 Number of students: 280 Examination date: 6 October 2014 Time of Examination: 13:00 13:50 Instructor: S. D Agostino No aids are permitted except for a non-programmable, non-graphics calculator. Communications with extraterrestrial and extra-dimensional entities during the test is strictly forbidden. Solve all problems in the space provided. Total number of marks: 16 SOLUTIONS 1. [1 point] A car is driving at a constant speed on a circular road. The force on a passenger in the car is (a) away from the centre of the circle, in the direction of the acceleration. (b) towards the centre of the circle, in the direction of the acceleration. (c) zero, by Newton s first law of motion, because the passenger is not accelerating. (d) zero, by Newton s first law of motion, because a body in motion tends to stay in motion. 2. [1 point] A soccer ball is kicked along the ground, and it rolls until it stops. The soccer ball (a) gradually slows down because the force of the kick gradually decreases as the ball rolls. (b) moves at a constant speed until the force of the kick dissipates and the ball suddenly slows down and stops. (c) slows down and stops because of Newton s third law of motion. (d) gradually slows down because friction opposes its motion. 3. [1 point] A 2-kg ball is thrown from the ground with an initial speed of 10 m/s at an initial angle of 30. A 1-kg ball is thrown from the ground with the same initial speed and the same initial angle. If there is no air resistance, (a) the 2-kg ball goes twice as far, because the gravitational force on it is twice as large. (b) the 1-kg ball goes twice as far, because according to Newton s second law its acceleration is twice as large. (c) the two balls have identical paths. (d) the 2-kg ball goes a bit farther, because it has more momentum. (e) the 1-kg ball goes a bit farther, because it is a little easier to throw.

2 For each of the statements below, state clearly whether the statement is true or false. If the statement is true, briefly explain why, with reference to the relevant laws of motion. If the statement is false, briefly explain why, with reference to the relevant laws of motion, and then correct the statement. 4. [2 points] Consider a passenger in a car driving forwards. When the driver slams on the brakes, the passenger is thrown forward by the force of the car s motion. Solution: False. Force is not something that can be carried along with an object, so there is no such thing as the force of the car s motion. By Newton s first law of motion, the passenger tends to continue moving forward at a constant speed when the car begins to slow down. Relative to the seat, the upper part of the passenger s body therefore moves forward a little relative to the lower part of the passengers s body. The lower part of the passenger s body doesn t move forward relative to the car because of the strong forces pushing it backwards from the seat belt and the seat. As the angle of the passenger s body increases, the lower part of the passenger s body exerts increasingly large forces on the upper part of the body (in the backwards direction) to slow the body down. As passengers, we feel like we are thrown forward because we have a self-centred perspective; we see the world relative to the reference frame of our bodies. But there is no force throwing us forward; on the contrary, the forces exerted on us by the seat and seat belt act in the backward direction, slowing us down. If there was a force acting on us in the forward direction, we would be catapulted out through the windshield at a non-healthy speed, because our initial velocity was high, and now there would be forward acceleration to boot. All you physics enthusiasts will learn in second year that as the car slows down, the reference frame of the passenger is non-inertial, and Newton s laws of motion don t apply in the simplest possible way in such reference frames. 5. [2 points] You use your car to push your friend s car forward, because its engine is broken. As you slow down together, the force that your car exerts on your friend s car is less than the force that your friend s car exerts on your car. Solution: False. By Newton s third law of motion, the force exerted by your car on your friend s car has the same magnitude as the force exerted by your friend s car on your car. The state of motion is irrelevant. 6. A baseball is thrown from the edge of a cliff at an angle of 32 above the horizontal with an initial speed of 19 m/s. The edge of the cliff is 16 m above the ground. (a) [1 point] Determine the time at which the ball reaches maximum altitude. (b) [1 point] Determine the maximum altitude. (c) [1 point] Determine the time at which the ball hits the ground. (d) [1 point] Determine how far from the cliff the ball hits the ground.

3 (e) [1 point] Determine the speed of the ball as it hits the ground. Solution: Start by sketching a diagram. On my diagram, I labelled the initial point of the ball s path 0, the point of the path where the ball reaches its peak height 1, and the point of the path where the ball hits the ground 2. I adopted the conventions that the time at which the ball is thrown is t 0 = 0, upwards is the positive direction, and the height of the ground is y 2 = 0. (a) To determine the time needed to reach the peak altitude, note that the vertical component of the ball s velocity at point 1 is zero. Thus, v y t = a y v y = a y t t = v y a y t 1 t 0 = v 1y v 0y 9.8 t 1 0 = 0 v 0 sin (19) sin 32 t 1 = 9.8 t 1 = 1.03 s Thus, it takes a little over one second for the ball to reach its peak altitude. (b) Use one of the y-displacement equations to determine the maximum altitude. For example, v 2 1y = v 2 0y + 2a y y y = v2 1y v 2 0y 2a y y 1 y 0 = 0 (v 0 sin 32 ) 2 2( 9.8) y 1 = 16 + (19 sin 32 ) 2 2(9.8) y 1 = m The peak altitude above the ground is 21.2 m. If you prefer to calculate the peak height above the top of the cliff, then the result is y 1 y 0 = = 5.17 m As a check, use the equation y = v 0y t a yt 2 1 to calculate the peak altitude; the result will be the same as we just calculated.

4 (c) To calculate the time needed to reach the ground, t 2, make use of the fact that y 2 = 0 when the ball hits the ground. Therefore, y = v 0y t a yt 2 2 y 2 y 0 = v 0 sin 32 t ( 9.8)t = 19 sin 32 t 2 4.9t 2 2 This is a quadratic equation for the time; place it in standard form and then use the quadratic formula: 4.9t sin 32 t 2 16 = 0 t 2 = 19 sin 32 ± (19 sin 32 ) 2 4(4.9)( 16) 2(4.9) t 2 = t 2 = 3.11 s The negative solution is rejected because it s not relevant for this problem. (All you physics enthusiasts might like to think about its interpretation, though.) Therefore, the ball hits the ground 3.11 s after it is thrown. (d) There is no acceleration in the x-direction; thus, the x-component of the ball s velocity is constant. Therefore, x = v 0x t 2 x 2 x 0 = v 0 cos 32 t 2 x 2 0 = (19) cos 32 (3.11) x 2 = m Thus, the ball hits the ground 50.0 m from the base of the cliff. (e) The x-component of the ball s velocity is constant; therefore, v 2x = v 0x = v 0 cos 32 = 19 cos 32 = m/s The y-component of the ball s impact velocity can be calculated as follows: v y t = a y v y = a y t v 2y v 0y = ( 9.8)t 2 v 2y = (19) sin (3.11) v 2y = Thus, using Pythagoras s theorem, the ball s impact speed is v 2 = v0x 2 + v0y 2 v 2 = v 2 = m/s

5 Thus, the ball s impact speed is 26.0 m/s. Notice that the impact speed is greater than the initial speed. Do you understand why this is so? 7. A ship is 49 km west of a straight shore that runs north-south. The ship moves at a constant speed of 19 km/h relative to the water at an angle of 23 north of east. There is a steady ocean current straight south with a speed of 12 km/h relative to the shore. (a) [2 point] Determine the velocity of the ship relative to the shore. (b) [1 point] Determine how long it takes for the ship to reach the shore. (c) [1 point] Determine how far south the ship has moved (relative to its starting position) when it reaches the shore. Solution: Sketch a diagram! (a) Let S represent the ship, let G represent the ground (the shore), and let W represent the water. The basic equation for relative velocity is The given information is and v SG = v SW + v W G v SW = (19 cos 23, 19 sin 23 ) v W G = (0, 12) Thus, using the basic equation for relative velocity, v SG = v SW + v W G v SG = (19 cos 23, 19 sin 23 ) + (0, 12) v SG = ( , ) + (0, 12) v SG = ( , ) If you prefer to quote the velocity of the ship in terms of speed and direction: and v SG = = km/h tan θ = ( ) θ = tan 1 = Thus, the velocity of the ship can also be quoted as 18.1 km/h with direction E14.7 S. (b) The time needed to reach the shore can be calculated by analyzing only the x- component of the motion: x = v 0x t t = x v 0x t = t = 2.80 h

6 The ship takes almost 3 hours to reach the shore. (c) The southward displacement can be determined by analyzing only the y-component of the motion: y = v 0y t y = ( )(2.80) y = km The ship moves nearly 13 km south relative to its starting position.