University:
Winston-Salem Forsyth County Schools
Course:
AP Physics I
Academic year:

2014
Views:

430

Pages:

3
Author:

Aracely Goodman

Physics Midterm Review Practice Problems 1. Larry fires a cannon ball that soars through the air for 5 seconds and lands 440 meters away. a. What is the horizontal velocity of the cannon? Δx = vΔt 440=v(5) 5 5 V =88 m/s b. What is the maximum height of the cannon? Δy = ½ a t2 + vi t Vertical velocity is zero at the top of the path. So, if we look at the “down” part of the motion, then vi=0. This is better here, because we have no information about the launch. Δy = ½ (-9.8) (2.5)2 + 0 Δy= -31 m Maximum Height= 31m c. What is the initial vertical velocity of the cannon? Δv=at vf-vi=at 0-vi = (-9.8)(2.5) -vi=-25m/s viy=25m/s d. What is the launch speed and angle for the cannon? θ = tan-1 (25/88) = 16 m/s V=? 25m/s a2 + b2 = c2 252 + 882 =c2 c=92 V=92m/s 88m/s 2. A 35N wagon is pushed across the floor with 400N of force at a 60o angle. The wagon starts from rest and speeds up to 32 m/s while covering 15 m of ground. a. What is the wagon’s acceleration? FN vf2=2aΔx + vi2 2 32 =2a(15) + 0 a=34 m/s b. How much friction acts on the wagon? ΣFx = ma 200N Ff -350 N 400N Fg =-35N Ff + x = ma Ff +200= (3.6)(34) Ff = 78 N Fx = 400 cos 60 = 200 N Fy = 400 sin 60 = -350 N c. What is the normal force acting on the wagon? ΣFy=ma 3. FN + Fg + y= 0 FN -35 -350 = 0 FN = 385N Use the graph below to answer the questions that follow. a. How does the object’s motion change at 40 seconds? The object begins to travel at a constant velocity. b. What is the acceleration of this object during the time interval from 0 to 40 seconds? Acceleration = slope = Δy = -200m/s = -5 m/s/s Δx 40s c. What is the object’s total displacement? Displacement = area After shading the area between the graph and the axes, the space can be broken down into a triangle and a rectangle. Δx = ½ bh + lw = ½ (40)(200) + (55)(50) = 6750 m 4. Two carts, each with a mass of 0.8 kg, sit at rest. They explode apart, sending one cart towards the left at 2.0 m/s. What is the speed of the second cart? Total Momentum before = Total Momentum after 0 = (0.8 kg) (-2.0 m/s) + (0.8 kg) v v = +2.0 m/s