PHY131 Ch 1 – 5 Exam Name

(30%) A quarterback throws a football toward you at a speed of 30.0 m/s and an angle of 30.0 degrees above the horizontal. But you are on top of a train (4.00 meters above ground level) that is 30 meters from the quarterback. As soon as the ball is released the train starts accelerating a rate of 2.00 m/s2 (moving away, with unknown initial velocity.) If you catch the ball, how far are you from the quarterback? / OR A kicker must kick a football from a point 50.0 m from the goal. Half the crowd hopes the ball will clear the crossbar, which is 3.00 meters high. When kicked, the ball leaves the ground with a speed of 20.0 m/s and an angle of 30.0 degrees above the horizontal. By how much does the ball clear or fall short of the crossbar?
y = ½ a t2 + vy t + yo
9/9 2/2 9/9
4 = ½-10t2 + 15 t + 0
t2 - 3t = -0.8
(t – 1.5)2 = -.8+1.52
t = 1.5 ± 1.2
t = 2.7 seconds / vx = 26 m/s (cos30*30)
vy = 15 m/s (sin30*30)
9/9 1/1
x = ½at2 + vx t + xo
xball = 26 t
xball = 70.2m
y = ½ a t2 + vy t + yo
3 = ½-10t2 + 10 t + 0
t2 - 2t = -0.6
t = 1 ± .63
t = 0.37 or 1.63 s / vx = 17.3 m/s
vy = 10 m/s
x = ½at2 + vx t + xo
xball = (17.3)1.63
xball = 28.3 m
Short by 21.7 m
Note: The question should have been … If you catch the ball, how fast was the train moving when the ball was released?
Then we need x = ½at2 + vx t + xo
xtrain = ½2t2 + vtrain t + 30
(40%) A 2.0 kg object moves in an xy plane according to x(t) = - 20 + 5 t - 4 t3 and
y(t) = 18 + 4 t - 10 t2 (SI Units).
At t = 2.0 s, what are (a) the magnitude and
(b) the angle (relative to the positive direction of the x axis) of the net force on the particle, and (c) what is the angle of the particle's direction of travel? / OR for ONLY 34%
Four blocks are connected by ropes. The masses are mA = 20 kg, mB = 30 kg, and mC = 40 kg, mD = 10 kg. When the system is released from rest,
(a) what is the tension in the cord connecting B and C /
(b) How far does B move in the first 0.50 sec?
x(t) = -20 + 5t - 4t3
vx(t) = 5 – 12t2
ax(t) = -24t 15/15 / y(t) = 18 + 4t - 10t2
vy(t) = 4 – 20t
ay(t) = -20
FNet = mT a
5/5 5/5 8/8 3/3
(mDg + mCg – mAg) = (mA+mB+mC+mD) a
(100 + 400 – 200) = (20+30+40+10) a
a = 3 m/s2
F = 2(-24t)i + 2(-20)j 4/4
F = -96 i - 40 j @ 2 sec
F = (962 + 402)^½ 4/4
F = 104 N
θ = tan-1(-40/-96) 4/4
θ = 202.6° (180+22.6) / v = (5-12t2) i + (4-20t) j
@ 2 sec 10/10
v = (-43) i + (-36) j
θ = tan-1(-36/-43) + 180°;
θ = 220° 3/3
1/1 2/2 6/6
FW = (mC + mD) (g – a)
FW = 50 (10 – 3)
FW = 350 N / x = ½ a t2
x = ½ 3 0.52 4/4
x = 0.38 meters
30% You see a mischievous monkey throwing rocks from a tree. When you are directly below him, you decide to throw your half eaten apple directly up toward the monkey at a speed of v. The monkey is ready and is 15 meters above your head. As you release your apple (at head level), he drops a rock. The rock impacts the apple at 10 meters. What is “v”?
Several students questioned this problem as being too simple and asked what speed the monkey threw the rock down. It would have been much better question if the Monkey threw the rock down at a speed of 3 m/s or similar. / yRock = ½ a t2 + vy t + yo
10 = ½-10t2 + 0 t + 15
t = 1 sec 14/14
yApple = ½ a t2 + vy t + yo
4/4 8/8 4/4
10 = ½-10 12 + v 1 + 0
v = 15 m/s