Problem#1
A person standing at the top of a hemispherical rock of radius R kicks a ball (initially at rest on the top of the rock) to give it horizontal velocity vi as in Figure 1. (a) What must be its minimum initial speed if the ball is never to hit the rock after it is kicked? (b) With this initial speed, how far from the base of the rock does the ball hit the ground? |
| Fig.1 |
Answer:
Measure heights above the level ground. The elevation yb of the ball follows
yf = yi + vyit – ½ gt2,
yb = R + 0 – ½ gt2,
Measure heights relative to the ground and call the ball’s starting position x = 0. The x position of the ball at any time t is just x = vit. The y position of the ball at any time t can be described by:
yb = R – ½ g(x/vi)2
For the last step, we made use of t = x/vi. Note that this is just the usual trajectory for projectile motion y(x) with the starting height yi = R added in. In order for the ball not to hit the rock, the parabola above must everywhere lie above the circle describing the rock. In other words, yb > yr at every x. We can write this as an inequality:
yb2 + x2 > R2
[R – ½ g(x/vi)2]2 + x2 > R2
R2 – (gx2R/vi2) + (x2g4/4vi4) + x2 > R2
(x2g4/4vi4) + x2 > (gx2R/vi2)
If this inequality is satisfied for x approaching zero, it will be true for all x. If the ball’s parabolic trajectory has large enough radius of curvature at the start, the ball will clear the whole rock:
1 > gR/vi2
vi > √(gR)
(b) With vi > √(gR) and yb = 0, we have
0 = R – (gx2/2gR)
x = R√2
The distance from the rock’s base is
x – R = R(√2 – 1)
Problem#2
A car is parked on a steep incline overlooking the ocean, where the incline makes an angle of 37.0° below the horizontal. The negligent driver leaves the car in neutral, and the parking brakes are defective. Starting from rest at t = 0, the car rolls down the incline with a constant acceleration of 4.00 m/s2, traveling 50.0 m to the edge of a vertical cliff. The cliff is 30.0 m above the ocean. Find (a) the speed of the car when it reaches the edge of the cliff and the time at which it arrives there, (b) the velocity of the car when it lands in the ocean, (c) the total time interval that the car is in motion, and (d) the position of the car when it lands in the ocean, relative to the base of the cliff.
Answer:
Measure heights above the level ground. The elevation yb of the ball follows
yf = yi + vyit – ½ gt2,
yb = R + 0 – ½ gt2,
Measure heights relative to the ground and call the ball’s starting position x = 0. The x position of the ball at any time t is just x = vit. The y position of the ball at any time t can be described by:
yb = R – ½ g(x/vi)2
For the last step, we made use of t = x/vi. Note that this is just the usual trajectory for projectile motion y(x) with the starting height yi = R added in. In order for the ball not to hit the rock, the parabola above must everywhere lie above the circle describing the rock. In other words, yb > yr at every x. We can write this as an inequality:
yb2 + x2 > R2
[R – ½ g(x/vi)2]2 + x2 > R2
R2 – (gx2R/vi2) + (x2g4/4vi4) + x2 > R2
(x2g4/4vi4) + x2 > (gx2R/vi2)
If this inequality is satisfied for x approaching zero, it will be true for all x. If the ball’s parabolic trajectory has large enough radius of curvature at the start, the ball will clear the whole rock:
1 > gR/vi2
vi > √(gR)
(b) With vi > √(gR) and yb = 0, we have
0 = R – (gx2/2gR)
x = R√2
The distance from the rock’s base is
x – R = R(√2 – 1)
Problem#2
A car is parked on a steep incline overlooking the ocean, where the incline makes an angle of 37.0° below the horizontal. The negligent driver leaves the car in neutral, and the parking brakes are defective. Starting from rest at t = 0, the car rolls down the incline with a constant acceleration of 4.00 m/s2, traveling 50.0 m to the edge of a vertical cliff. The cliff is 30.0 m above the ocean. Find (a) the speed of the car when it reaches the edge of the cliff and the time at which it arrives there, (b) the velocity of the car when it lands in the ocean, (c) the total time interval that the car is in motion, and (d) the position of the car when it lands in the ocean, relative to the base of the cliff.
Answer:
 |
| Fig.2 |
(a) While on the incline
vB2 = vA2 + 2adAB
vB2 = 0 + 2(4 m/s2)(50 m)
vB = 20 m/s
and
tAB = (vB - vA)/a = (20 m/s – 0)/(4 m/s2)
tAB = 5 s
(b) Initial free-flight conditions give us
vxB = vi cos 370 = (20 m/s) cos 370 = 16.0 m/s
vyB = vi sin 370 = –(20 m/s) sin 370 = –12 m/s
vxC = vxA = 16.0 m/s and
vyC = –√[2(–g)∆y + vyB2]
vyC = –√[2(–9.80 m/s2)(–30 m) + (–12 m)2] = –27.1 m/s
so that
vc = √[vxC2 + vyC2] = √[(16.0 m/s)2 + (–27.1 m/s)2] = 31.5 m/s and
α = tan-1(vyC/vxC) = tan-1((–27.1/16.0) = 59.40 below the horizontal
(c) vyC = vyB – gtBC
–27.1 m/s = –12 m/s – (9.80 m/s2)tBC
tBC = 1.53 s
then t = tAB + tBC = 6.53 s
(d) the position of the car when it lands in the ocean, relative to the base of the cliff is
∆x = vxBt = (16.0 m/s)(1.53 s) = 24.5 m
Problem#3
A truck loaded with cannonball watermelons stops suddenly to avoid running over the edge of a washed-out bridge (Fig. 3). The quick stop causes a number of melons to fly off the truck. One melon rolls over the edge with an initial speed vi = 10.0 m/s in the horizontal direction. A cross-section of the bank has the shape of the bottom half of a parabola with its vertex at the edge of the road, and with the equation y2 = 16x, where x and y are measured in meters. What are the x and y coordinates of the melon when it splatters on the bank?
vB2 = vA2 + 2adAB
vB2 = 0 + 2(4 m/s2)(50 m)
vB = 20 m/s
and
tAB = (vB - vA)/a = (20 m/s – 0)/(4 m/s2)
tAB = 5 s
(b) Initial free-flight conditions give us
vxB = vi cos 370 = (20 m/s) cos 370 = 16.0 m/s
vyB = vi sin 370 = –(20 m/s) sin 370 = –12 m/s
vxC = vxA = 16.0 m/s and
vyC = –√[2(–g)∆y + vyB2]
vyC = –√[2(–9.80 m/s2)(–30 m) + (–12 m)2] = –27.1 m/s
so that
vc = √[vxC2 + vyC2] = √[(16.0 m/s)2 + (–27.1 m/s)2] = 31.5 m/s and
α = tan-1(vyC/vxC) = tan-1((–27.1/16.0) = 59.40 below the horizontal
(c) vyC = vyB – gtBC
–27.1 m/s = –12 m/s – (9.80 m/s2)tBC
tBC = 1.53 s
then t = tAB + tBC = 6.53 s
(d) the position of the car when it lands in the ocean, relative to the base of the cliff is
∆x = vxBt = (16.0 m/s)(1.53 s) = 24.5 m
Problem#3
A truck loaded with cannonball watermelons stops suddenly to avoid running over the edge of a washed-out bridge (Fig. 3). The quick stop causes a number of melons to fly off the truck. One melon rolls over the edge with an initial speed vi = 10.0 m/s in the horizontal direction. A cross-section of the bank has the shape of the bottom half of a parabola with its vertex at the edge of the road, and with the equation y2 = 16x, where x and y are measured in meters. What are the x and y coordinates of the melon when it splatters on the bank?
 |
| Fig.3 |
Answer:
Equations of motion
x-axis: ∆x = x = vxit
y-axis: yf = yi + vyit – ½ gt2, with vyi = 0, then
y = – ½ gt2
from (*),
y = – ½ g(x/vi)2
Equate y from the bank equation to y from the equations of motion:
y2 = [– ½ g(x/vi)2]2
16x = [– ½ g(x/vi)2]2
g2x4/(4vi4) – 16x = 0
x[g2x3/(4vi4) – 16] = 0
x = 0 or
g2x3/(4vi4) – 16= 0
g2x3/(4vi4) = 16
x3 = 64vi4/g2
x = 4[(10 m/s)4/(9.80 m/s2)]1/3 = 18.8 m.
Also,
y = – ½ g(x/vi)2
y = – ½ (9.80 m/s2)(18.8 m/10 m/s)2 = –17.3 m
Equations of motion
x-axis: ∆x = x = vxit
y-axis: yf = yi + vyit – ½ gt2, with vyi = 0, then
y = – ½ gt2
from (*),
y = – ½ g(x/vi)2
Equate y from the bank equation to y from the equations of motion:
y2 = [– ½ g(x/vi)2]2
16x = [– ½ g(x/vi)2]2
g2x4/(4vi4) – 16x = 0
x[g2x3/(4vi4) – 16] = 0
x = 0 or
g2x3/(4vi4) – 16= 0
g2x3/(4vi4) = 16
x3 = 64vi4/g2
x = 4[(10 m/s)4/(9.80 m/s2)]1/3 = 18.8 m.
Also,
y = – ½ g(x/vi)2
y = – ½ (9.80 m/s2)(18.8 m/10 m/s)2 = –17.3 m
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