Problem#1
Catapulting mushrooms. Certain mushrooms launch their spores by a catapult mechanism. As water condenses from the air onto a spore that is attached to the mushroom, a drop grows on one side of the spore and a film grows on the other side. The spore is bent over by the drop’s weight, but when the film reaches the drop, the drop’s water suddenly spreads into the film and the spore springs upward so rapidly that it is slung off into the air. Typically, the spore reaches a speed of 1.6 m/s in a 5.0 µm launch; its speed is then reduced to zero in 1.0 mm by the air. Using that data and assuming constant accelerations, find the acceleration in terms of g during (a) the launch and (b) the speed reduction.Answer:
In this problem we are given the initial and final speeds, and the displacement, and are asked to find the acceleration. We use the constant-acceleration equation given in Eq.
v2 = v02 + 2a(x – x0)
(a) Given that v0 = 0 , v =1.6 m/s, and Δx = 5.0 μm, the acceleration of the spores during the launch is
a = (v2 – v02)/2∆x = (1.6 m/s)2/[2(5 x 10-6 m)] = 2.56 x 105 m/s2 = 2.6 x 104g
(b) During the speed-reduction stage, the acceleration is
a = (v2 – v02)/2∆x = [0 – (1.6 m/s)2]/[2(1 x 10-3 m)] = –1.28 x 103 m/s2 = –1.3 x 102g
The negative sign means that the spores are decelerating.
Problem#2
A muon (an elementary particle) enters a region with a speed of 5.00 x 106 m/s and then is slowed at the rate of 1.25 x 1014 m/s2. (a) How far does the muon take to stop? (b) Graph x versus t and v versus t for the muon.
Answer:
(a) Setting v = 0 and x0 = 0 in v2 = v02 + 2a(x – x0)
we find
x = –½ v02/a = – ½ (5.00 x 106 m/s)2/(–1.25 x 1014 m/s2) = 0.100 m
Since the muon is slowing, the initial velocity and the acceleration must have opposite signs.
(b) Below are the time plots of the position x (Fig.1a) and velocity v (Fig.1b) of the muon from the moment it enters the field to the time it stops. The computation in part (a) made no reference to t, so that other equations v = v0 + at and x = x0 + v0t + ½ at2 are used in making these plots.
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| Fig.1 |
On a dry road, a car with good tires may be able to brake with a constant deceleration of 4.92 m/s2. (a) How long does such a car, initially traveling at 24.6 m/s, take to stop? (b) How far does it travel in this time? (c) Graph x versus t and v versus t for the deceleration.
Answer:
We take +x in the direction of motion, so v0 = +24.6 m/s and a = – 4.92 m/s2 . We also take x0 = 0.
(a) The time to come to a halt is found using Eq v = v0 + at, then
0 = (24.6 m/s) + (– 4.92 m/s2)t
t = 5.00 s
(b) we use the eq. v2 = v02 + 2a∆x
0 = (24.6 m/s)2 + 2(– 4.92 m/s2)∆x
∆x = 61.5 m
(c) Using these results, we plot v0t + ½ at2 (the x graph (Fig.2a), shown next, on the left) and v0 + at (the v graph (Fig.2b), on the right) over 0 ≤ t ≤ 5 s, with SI units understood.
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| Fig.2 |
The brakes on your car can slow you at a rate of 5.2 m/s2. (a) If you are going 137 km/h and suddenly see a state trooper, what is the minimum time in which you can get your car under the 90 km/h speed limit? (The answer reveals the futility of braking to keep your high speed from being detected with a radar or laser gun.) (b) Graph x versus t and v versus t for such a slowing.
Answer:
We choose the positive direction to be that of the initial velocity of the car (implying that a < 0 since it is slowing down). We assume the acceleration is constant.
(a) Substituting v0 = 137 km/h = 38.1 m/s, v = 90 km/h = 25 m/s, and a = –5.2 m/s2 into
v = v0 + at
we obtain
25 m/s = 38.1 m/s + (–5.2 m/s2)t
t = 2.5 s
(b) We take the car to be at x = 0 when the brakes are applied (at time t = 0). Thus, the coordinate of the car as a function of time is given by
x = x0 + v0t + ½ at2
x = 0 + (38 m/s)t + ½ (–5.2 m/s2)t2
in SI units. This function is plotted from t = 0 to t = 2.5 s on the graph to the right. We have not shown the v-vs-t graph (Fig.3) here; it is a descending straight line from v0 to v.
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| Fig.3 |
Problem#5
A world’s land speed record was set by Colonel John P. Stapp when in March 1954 he rode a rocket-propelled sled that moved along a track at 1020 km/h. He and the sled were brought to a stop in 1.4 s. (See Fig. 4) In terms of g, what acceleration did he experience while stopping?
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| Fig.4 |
Answer:
The acceleration is found from a = ∆v/∆t,
With v = (1020 km/h)(1000 m/1 km)(1 h/3600 s) = 283.33 m/s
Then
a = (283.33 m/s)/(1.4 s) = 202.4 m/s2
In terms of the gravitational acceleration g, this is expressed as a multiple of 9.8 m/s2 as follows:
a = [(202.4 m/s2)/(9.81 m/s2)] = 21g
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