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Physics 6A Practice Final fall 2010 solutions

# Physics 6A Practice Final fall 2010 solutions - Physics 6A...

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Physics 6A Practice Final (Fall 2010) solutions Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

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1) A helicopter traveling upwards at 121 m/s drops a package from a height of 500m above the ground. Assuming free-fall, how long does it take to hit the ground? This is a straightforward kinematics problem. We can take up to be positive, so the initial velocity is v 0 =121 m/s and the acceleration will be –g. Initial height is y 0 =500m. Using the basic formula for height in free fall: t ) 8 . 9 ( t ) 121 ( m 500 0 gt t v y y 2 s m 2 1 s m 2 2 1 0 0 2 + = + = Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB sec 28 t ) 9 . 4 ( 2 ) 500 )( 9 . 4 ( 4 ) 121 ( 121 t 0 500 t 121 t 9 . 4 2 2 = ± = = We can use the quadratic formula and take the positive answer. This is answer c .
This is a projectile problem with a horizontal initial velocity. Here are the initial and final values that we know: g a 0 v 5 . 3 v m 0 y m 5 . 1 y s m y , 0 s m x , 0 0 = = = = = 2) A person skateboarding with a constant speed of 3.5 m/s releases a ball from a height of 1.5m above the ground. Find the speed of the ball as it hits the ground. When the ball hits the ground we will have the same x-component of velocity, but the y-component will have increased. ( ) 0 2 y , 0 2 y y y g 2 v v = y Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB s m y s m 2 y 42 . 5 v ) m 5 . 1 0 )( 8 . 9 ( 2 0 v 2 = = We need to use the Pythagorean theorem to find the magnitude of the velocity: s m 2 2 2 y 2 x 45 . 6 ) 42 . 5 ( ) 5 . 3 ( v v v v = + = + = Answer c

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s m s m 0 0 y , 0 s m s m 0 0 x , 0 47 . 11 ) 35 sin( ) 20 ( ) sin( v v 38 . 16 ) 35 cos( ) 20 ( ) cos( v v = = θ = = = θ = o o We will need to find the components of the initial velocity: 3) A projectile is launched from the origin with an initial speed of 20.0m/s at an angle of 35.0° above the horizontal. Find the maximum height attained by the projectile. The vertical component of velocity will be 0 when it reaches the highest point, so we can use a kinematics formula to find the maximum height: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB ( ) m 71 . 6 y ) 0 y )( 8 . 9 ( 2 ) 47 . 11 ( 0 y y g 2 v v max max s m 2 s m 0 2 y , 0 2 y 2 = = = Answer b
4) You are standing on a scale in an elevator. When the elevator is at rest the scale reads 750 N. You press the button for the top floor and the elevator begins to accelerate upward at a constant rate. If the scale now reads 850 N what is the acceleration of the elevator? F scale Draw a free-body diagram for the person. The original reading on the scale is the person’s weight. Divide by 9.8 to get the mass. Write down Newton’s 2 nd law: ma F = Σ mg 2 s m scale 3 . 1 a a ) kg 5 . 76 ( N 750 N 850 ma mg F = = = Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Answer a

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5) Two blocks are connected by a string and are pulled vertically upward by a force F = 90N applied to the upper block. Find the tension in the string connecting the two blocks.
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