ssm_ch09 - Chapter 9 Student Solutions Manual 15. We need...

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Chapter 9 – Student Solutions Manual 15. We need to find the coordinates of the point where the shell explodes and the velocity of the fragment that does not fall straight down. The coordinate origin is at the firing point, the + x axis is rightward, and the + y direction is upward. The y component of the velocity is given by v = v 0 y gt and this is zero at time t = v 0 y / g = ( v 0 / g ) sin θ 0 , where v 0 is the initial speed and 0 is the firing angle. The coordinates of the highest point on the trajectory are () 2 2 0 00 0 2 20 m/s cos sin cos sin 60 cos60 17.7 m 9.8 m/s x v xv tv t g θθ == = = ° ° = and yv t g t v g y =− = = = ° 0 2 0 2 2 0 2 2 1 2 1 2 1 2 20 98 60 153 sin . . m/s m. 2 b g Since no horizontal forces act, the horizontal component of the momentum is conserved. Since one fragment has a velocity of zero after the explosion, the momentum of the other equals the momentum of the shell before the explosion. At the highest point the velocity of the shell is v 0 cos 0 , in the positive x direction. Let M be the mass of the shell and let V 0 be the velocity of the fragment. Then Mv 0 cos 0 = MV 0 /2, since the mass of the fragment is M /2. This means Vv 00 0 2 2 20 60 20 = ° cos cos m/s. b g This information is used in the form of initial conditions for a projectile motion problem to determine where the fragment lands. Resetting our clock, we now analyze a projectile launched horizontally at time t = 0 with a speed of 20 m/s from a location having coordinates x 0 = 17.7 m, y 0 = 15.3 m. Its y coordinate is given by yy g t 0 1 2 2 , and when it lands this is zero. The time of landing is ty = 2 0 / g and the x coordinate of the landing point is xxV txV y g =+ = + = 0 2 17 7 2153 53 . . . m 20 m / s m 2 b g b g 23. The initial direction of motion is in the +x direction. The magnitude of the average force F avg is given by 3 2 32.4 N s 1.20 10 N 2.70 10 s avg J F t = × Δ× The force is in the negative direction. Using the linear momentum-impulse theorem stated in Eq. 9-31, we have F avg Δ t = mv f mv i .
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where m is the mass, v i the initial velocity, and v f the final velocity of the ball. Thus, () ( ) ( )( ) 3 avg 0.40kg 14m s 1200 N 27 10 s 67m s. 0.40kg i f mv F t v m −× −Δ == = (a) The final speed of the ball is || f v = 67 m/s. (b) The negative sign indicates that the velocity is in the – x direction, which is opposite to the initial direction of travel. (c) From the above, the average magnitude of the force is . 3 avg 1.20 10 N F (d) The direction of the impulse on the ball is – x , same as the applied force. 35. (a) We take the force to be in the positive direction, at least for earlier times. Then the impulse is 33 3 3.0 10 69 00 3.0 62 93 0 (6.0 10 ) (2.0 10 ) 11 23 9.0N s. JF d t t t tt −− ×× × 2 d t × × ⎡⎤ × ⎢⎥ ⎣⎦ =⋅ ∫∫ (b) Since J = F avg Δ t , we find F J t avg 3 9.0 N s 3.0 10 s 3.0 10 N. Δ = × 3 (c) To find the time at which the maximum force occurs, we set the derivative of F with respect to time equal to zero – and solve for t . The result is t = 1.5 × 10 –3 s. At that time the force is F max 6.0 10 10 2.0 10 10 4.5 10 N. × × = × ch 15 2 ..
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ssm_ch09 - Chapter 9 Student Solutions Manual 15. We need...

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