Problem Set: Fall #1 - Solutions

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Felix Franklin
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1 Problem Set: Fall #1 - Solutions 1. (a) The car stops speedin up in the neative direction and beins deceleratin, probably brakin. (b) Calculate the averae velocity over each time interval. v av0 v 0 + v (0 m/s) + ( 0 m/s) 10 m/s v av 5 v + v 5 ( 0 m/s) + (0 m/s) 10 m/s The averae velocities are equal. (c) The displacement is the area under the curve. x 1 (5 s)( 0 m/s) + 1 (3 s)(0 m/s) 0 m (d) The slope of the iven raph represents the acceleration. Here is the acceleration of the car as a function of time: Acceleration (m/s ) Time (s) (e) The area of the iven raph represents the displacement. Here is the position of the car as a function of time: 1

2 0 Time (s) Position (m) (a) Use the constant acceleration equations. v 1f v 1i + a 1 t 1end t 1end v 1 f v 1i a 1 0 ( (90 km/h) ( 1 h 3600 s ) (1000 m 1 km )) m/s 1.5 s (b) Aain, use the constant acceleration equations. t 1end x 1end v 1 i + v 1end ( ) 5 m/s + 0 (1.5 s) m (c) There are two possible issues here. Check that the trailin car doesn t travel past where the lead car stops, and also make sure it doesn t hit the lead car at any time while they are stoppin. Here is how to use the first condition, d representin the extra distance. x end < x 1end + d v end v i + a x end 0 v i + a x end a < < v i x end v i ( x 1end + d) (35 m/s) (156.5 m + 45 m) < 3.04 m/s The other possibility is easiest to check usin the viewpoint of one of the cars. This solution uses the viewpoint of the lead car. The trailin car must have no relative velocity when it catches the

3 lead car; then if this acceleration s manitude is reater than that of the lead car there will not be a collision. v f v i + a x 0 v i + a x a x v i a v i x < v i < d (35 m/s 5m/s) (45 m) < 1.11 m/s Now convert out of the accelerated reference frame. a a + a 1 < ( 1.11 m/s ) + ( m/s ) < 3.11 m/s Notice usin only the first method would have left us with a collision. 3. (a) Use the horizontal components to find the time. t x v ix x v i cos40 50 m (55 m/s)cos s (b) Find the heiht of the cannonball at this time. y v iy t + 1 a( t) v i sin 40 t 1 ( t) xtan 40 1 ( x ) v i cos40 (50 m)(tan 40 ) 1 ( ( m ) m/s ) (55 m/s)cos m Since this is less than 40 m and more than 0, the cannonball strikes the wall 37.3 m above the round. (c) Use the horizontal components to find the time it takes to reach the wall. t x v ix x v i cosθ 3

4 Use this time with the vertical components to set the ball s heiht above the wall. y < v iy t + 1 a( t) < v i sin θ t 1 ( t) ( ) x < v i sin θ 1 ( x v i cosθ v i cosθ < xtan θ ( x) (1 + tan θ) ( ( x) v i tan θ v i ) tan θ ( x)tanθ + ) ) ( y + ( x) < 0 This is a quadratic for tanθ. ( )( ) x ± ( x) ( x) vi y + ( x) vi (50 m) ± ( x) v i (50 m) v i ( )( ) (9.8 m/s )(50 m) (55 m/s) (40 m) + (9.8 m/s )(50 m) (55 m/s) (9.8 m/s )(50 m) (55 m/s) < tanθ < < θ < (a) At the maximum heiht the vertical velocity component is 0. v y f v y i + a y y 0 v sin θ H H v sin θ H v sin θ (b) When it reaches its rane its vertical displacement component is 0. Use this to find the time required. y v yi t + 1 a y( t) 0 v sin θ t 1 ( t) v sin θ t t v sin θ Use this time to find the rane from the horizontal components. x v xi t 4

5 R v cosθ t ( ) v v cosθ sin θ v sin(θ) (c) The rane is maximized when sin(θ) is maximized. sin(θ) 1 θ 90 θ 45 (d) Find the times when the projectile passes throuh heiht h. y v yi t + 1 a( t) h v sin θt h 1 t h ( ) t h + ( v sin θ)t h + (h) 0 Use the quadratic formula to find the two times. t h v sin θ ± (v sin θ) h Find the difference between these times t h v sinθ + (v sinθ) h v sin θ h v sin θ (v sin θ) h 5. (a) The vertical component of the displacement is 0 over each parabola, just as in 4b. Use this to find the time taken for each parabola. y v yi t + 1 a y( t) 0 v sin φ t 1 ( t) v sinφ t t v sin φ Use this time to find the rane of each parabola. x v xi t + 1 a x( t) v cosφ t + 0 ( ) v v cosφ sin φ 5

6 v sin(φ) For the no-bounce throw: D v. For the first part of the one-bounce throw: d 1 v sin(θ). For the second part of the one-bounce throw: d 16v 100 sin(θ). Set the total ranes equal and solve for θ. d 1 + d D v 116v v v sin(θ) + sin(θ) 100 sin(θ) v sin(θ) Notice that there are two solutions. However, the steeper anled solution requires the ball to take loner to arrive, so it is not the solution with which we are concerned. θ 1 sin (b) Use the equation for time found in part a. v For the no-bounce throw: T. For the first part of the one-bounce throw: t 1 v sinθ. For the first part of the two-bounce throw: t 8v sin θ. 10 Ratio t 1 + t T v sinθ + 8v 10 sinθ v 8 10 sin θ 8 10 sin

Linear Motion. Miroslav Mihaylov. February 13, 2014

Linear Motion. Miroslav Mihaylov. February 13, 2014 Linear Motion Miroslav Mihaylov February 13, 2014 1 Vector components Vector A has manitude A and direction θ with respect to the horizontal. On Fiure 1 we chose the eastbound as a positive x direction