Physics 201 Homework 1 Jan 9, 2013 1. (a) What is the magnitude of the average acceleration of a skier who, starting (a) 1.6 m/s 2 ; (b) 20 meters from rest, reaches a speed of 8.0 m/s when going down a slope from 5.0 seconds? (b) How far does the skier travel in this time? (a) The quantities in play are: a =? v = 8.0 t = 5.0 The equation to use is v = v 0 + at. Thus, (b) Now the quantities are: (8.0) = (0) + (a)(5.0) = a = 1.6 x =? v = 8.0 t = 5.0 The equation to use is x = 1 2 (v 0 + v)(t). Thus, (x) = 1 2 (0 + 8.0)(5.0) = x = 20 2. A car is traveling at a constant speed of 33 m/s on a highway. At the 0.87 m/s 2 instant this car passes an entrance ramp, a second car enters the highway from the ramp. The second car starts from rest and has a constant acceleration. What acceleration must it maintain, so that the two cars meet for the first time at the next exit, which is 2.5 km away? Notice that of the two cars, the first moves with constant velocity. So the equation that governs his motion is x = vt. We know its velocity: 33 m/s. We also know its displacement: 2500 meters. So, we can calculate the time interval associated with the problem: (2500) = (33)(t) = t = 75.758 That s about as much as we can do with the first car. involving the second car are: a =? x = 2500 t = 75.758 Now, the quantities By starting with the simpler motion, we have enough data to solve the more complex one. The equation to use is x = v 0 t + 1 2 at2. Thus, (2500) = (0)(75.758) + 1 2 (a)(75.758)2 = a = 0.8712 3. A dynamite blast at a quarry launches a chunk of rock straight upward, and (a) 35 m/s; (b) 14 m/s 1
2.0 seconds later it is rising at a speed of 15 m/s. Assuming air resistance has no effect on the rock, calculate its speed (a) at launch and (b) 5.0 seconds after launch. (a) This is a free-fall problem since the force involved is gravity. This implies that m/s 2. Notice the negative sign. Listing the quantities in the problem: t = 2.0 v = +15 v 0 =? I ve called out the positive sign on v because we know that the chunk is still rising at this point. These four quantities point to using v = v 0 + at, so... (+15) = (v 0 ) + ( 9.80)(2.0) = v 0 = +34.600 The final positive sign indicates that the initial velocity is up which we knew, but it s helpful to use the signs to confirm the algebra. (b) Now that we know the initial launch velocity, we can solve this part using t = 5.0 v 0 = +34.600 v =? The same equation, v = v 0 + at, is useful: (v) = (+34.600) + ( 9.80)(5.0) = v = 14.400 In this case, the final velocity is negative. This indicates that the chunk of rock is now falling down. By the way, there is another (similar) way to do this problem. We know the velocity after 2.0 seconds is +15 m/s. We could treat this as our initial velocity. The duration would then be 3.0 seconds. Thus, t = 3.0 v 0 = +15 v =? We are still using the same equation, v = v 0 + at, so... (v) = (+15) + ( 9.80)(3.0) = v = 14.400 Which gives the same answer. This approach is a bit more tricky but has the advantage that it does not rely on my answer from part (a). This protects the second part from any error I may have made in the first part. In either case, we are asked for the speed instead of the velocity, so drop the sign. 4. At the beginning of a basketball game, a referee tosses the ball straight up 0.47 seconds with a speed of 4.6 m/s. A player cannot touch the ball until after it reaches its maximum height and begins to fall down. What is the minimum time that a player must wait before touching the ball? 2
Start by listing the quantities I am given: v 0 = 4.6 v = 0 In this problem there are two pieces of implicit data. First, since the problem is a free-fall problem, a = -9.80. The second is the term maximum height. Although the ball is always accelerating, at the top of its trajectory the instantaneous velocity is zero. At that moment, the ball is at rest. But even though it is at rest, the ball is still under the influence of gravity its acceleration is still the constant -9.80 m/s 2. In other words, the velocity of the ball goes to zero and continues to change. But we are not interested in the trajectory after peak because we want to know the minimum time the player needs to wait. This occurs just when the peak is reached. Given the quantities listed above, the equation to use is v = v 0 + at. Thus, (0) = (4.6) + ( 9.80)(t) = t = 0.46939 5. While standing on a bridge 15.0 meters above the ground, you drop a stone 11.3 m/s from rest. When the stone has fallen 3.20 meters, you throw a second stone straight down. What initial velocity must you give the second stone if they are both to reach the ground at the same instant? Both stones are in free-fall, so neither are in constant velocity motion. Let s just start by listing the quantities for the first stone: x = 15.0 I chose to include t because the question uses the words at the same instant. Of the two possible extra quantities (t or v) this seems like the best to pick. We can always go back and figure out v later if we need it. So, the equation to use is x = v 0 t + 1 2 at2. Thus, ( 15.0) = (0)(t) + 1 2 ( 9.80)(t)2 = t = 1.7496 Well, now what? What about the second stone? It s initial velocity is definitely not zero in fact this is the quantity we are seeking. We also know that we need to wait until the first stone has fallen 3.20 meters before we throw the second one. How much time do we wait? We have... And... x = 3.20 ( 3.20) = (0)(t) + 1 2 ( 9.80)(t)2 = t = 0.80812 So, we must wait 0.81 seconds. Now, the total amount of time the first stone is falling is 1.75 seconds, so the second stone only has 0.94 seconds to cover the 3.20 meters to the ground. Let s calculate this duration a bit more precisely: t = 1.7496 0.80812 = 0.94148 3
This is how long the second stone is in free-fall. What else do we know about it? t = 0.94148 x = 15.0 v 0 =? So, we have enough data now to answer the original question. The equation to use is x = v 0 t + 1 2 at2. Thus, ( 15.0) = (v 0 )(0.94148) + 1 2 ( 9.80)(0.94148)2 = v 0 = 11.319 Notice the negative sign. Clearly we need to be throwing the stone down. This is about a 25 mph throw. 6. A major-league pitcher can throw a baseball in excess of 41.0 m/s. If a ball is 0.84 meters thrown horizontally at this speed, how much will it drop by the time it reaches a catcher who is 17.0 meters away from the point of release? Let s begin with the horizontal since we know both the velocity and displacement. The horizontal motion is governed by x = vt, so we can calculate the time involved: (17.0) = (41.0)(t) = t = 0.41463 Now let s focus on the vertical motion. We have: t = 0.41463 y =? The equation to use is y = v 0 t + 1 2 at2. Thus, (y) = (0)(0.41463) + 1 2 ( 9.80)(0.41463) = y = 0.84242 7. A rocket is fired at a speed of 75.0 m/s from ground level, at an angle of 33.2 meters 60.0 above the horizontal. The rocket is fired toward an 11.0-meter wall, which is located 27.0 meters away. The rocket attains its launch speed in a negligibly short period of time, after which its engines shut down and the rocket coasts. By how much does the rocket clear the top of the wall? Okay, let s break this initial velocity into its components: v 0x = v 0 cos θ = (75.0) cos(60.0 ) = 37.500 v 0y = v 0 sin θ = (75.0) sin(60.0 ) = 64.952 In this case, let s start with the horizontal motion since we know the horizontal displacement. The horizontal equation is x = vt, so (27.0) = (37.500)(t) = t = 0.72000 That s it for the horizontal. The vertical data we have is: v 0 = 64.952 t = 0.72000 y =? 4
The equation to use is y = v 0 t + 1 2 at2. Thus, (y) = (64.952)(0.72000) + 1 2 ( 9.80)(0.72000)2 = y = 44.225 Since we are interested in how much this projectile clears an 11.0 meter wall, we need to subtract: 44.225 11.0 = 33.225 8. A soccer player kicks the ball toward a goal that is 16.8 meters in front of him. 14.7 m/s The ball leaves his foot at a speed of 16.0 m/s and an angle of 28.0 above the ground. Find the speed of the ball when the goalie catches it in front of the net. Since we know about the initial velocity, let s break it into its components: v 0x = v 0 cos θ = (16.0) cos(28.0 ) = 14.127 v 0y = v 0 sin θ = (16.0) sin(28.0 ) = 7.5115 We know that the horizontal component won t change, so it remains to determine the final vertical velocity. The only other thing we know is the horizontal displacement. We can t use the range formula because the vertical displacement may not be zero. But we can use this information to determine the time of flight using x = vt: (16.8) = (14.127)(t) = t = 1.1892 In the vertical we know have: v 0 = 7.5115 t = 1.1892 v =? The equation that links these quantities is v = v 0 + at, so... (v) = (7.5115) + ( 9.80)(1.1892) = v = 4.1427 This is the vertical component. Now we can calculate the magnitude of the velocity: mag v = vx 2 + vy 2 = (14.127) 2 + ( 4.1427) 2 = (14.127) 2 + ( 4.1427) 2 = 14.722 9. A projectile is launched from ground level at an angle of 12.0 above the 27.2 horizontal. It returns to ground level. To what value should the launch angle be adjusted, without changing the launch speed, so that the range doubles? Even though this is a double star problem, it s pretty easy if you use the range equation: So the first launch is summarized by R = R = v2 g sin 2θ v2 9.80 sin 2(12 ) = R = (0.041504)(v 2 ) 5
Now, we know that the launch speed will not change, so the second launch is summarized by 2R = sin 2θ 9.80 We can combine these two equations by substituting the first into the second: v2 (2)(0.041504)(v 2 ) = v2 sin 2θ 9.80 Notice that we can cancel the v 2 from both sides now. This leaves (0.08301) = 1 sin 2θ 9.80 Multiply both sides by 9.80 and solve for θ... (0.81347) = sin 2θ = θ = 27.218 10. A golfer standing on a fairway, hits a shot to a green that is elevated 5.50 5.17 seconds meters above the point where she is standing. If the ball leaves her club with a velocity of 46.0 m/s at an angle of 35.0 above the ground, find the time that the ball is in the air before it hits the green. The components of the initial velocity are: v 0x = v 0 cos θ = (46.0) cos 35.0 = 37.681 v 0y = v 0 sin θ = (46.0) sin 35.0 = 26.385 We don t know enough to use the horizontal equation x = vt, so let s focus on the vertical. The quantities involved are: v 0 = 26.385 y = 5.50 It looks like we can solve this problem with just the vertical motion. The equation to use is y = v 0 t + 1 2 at2. Thus, (5.50) = (26.385)(t) + 1 2 ( 9.80)(t)2 This is a quadratic equation. Rewritten in standard form, we have: ( 4.90)t 2 + (26.385)t + ( 5.50) = 0 Plugging this into the quadratic equation yields: t = b ± b 2 4ac 2a = (26.385) ± (26.385) 2 (4)( 4.90)( 5.50) (2)( 4.90) = 26.385 ± 24.256 9.80 6
= 0.21721 or 5.1675 The first answer corresponds to the moment just after the ball has left the club and is 5.50 meters up the ball is rising at this point. But we are interested in the moment after the ball has reached its peak and falls back to this height. So we want t = 5.1675. 7