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Vladimir [108]
4 years ago
9

You have been hired to design a spring-launched roller coaster thatwill carry two passengers per car. The car goes up a 10-m hig

hhill, then descends 15m to the track's lowest point. You'vedetermined that the spring can be compressed a maximum o 2.0m andthat a loaded car will have a maximum mass of 400kg. For safetyreasons, the spring constant should be 10% larger than the minimumneeded for the car to just make it over the top.a) what spring constant (k) should you specify ?b) What is the maximum speed of a 350 kg car if the spring iscompressed the full amount.
Physics
1 answer:
snow_tiger [21]4 years ago
3 0

Answer:

a) 21.6 KN/m

b)v = 18.56 m/s

Explanation:

For this problem, the spring should be strong enough to get the car just over the top so:

mgh = (1/2)kx^2

(400)*(9.8)*(10) = (0.5)*(k)*(2.0^2)

k = 39200/2.645

k = 19600 N/m

a) K increased by Increase by 10%, so:

k = 19600 * 1.10 = 21560 N/m = 21.6KN/m---->specified spring constant

b) Now, Maximum speed of a 350 kg car will be at the bottom of the roller coaster:

therefore, (1/2)mv^2 + mg(Y_1-Y_2) = (1/2)kx^2

(0.5)*(350)*(v^2) + (350)(9.8)(10-15) =(0.5)*(21560)*(2.0^2)

v^2 = 344.4

v = 18.56 m/s

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the 200 g baseball has a horizontal velocity of 30 m/s when it is struck by the bat, B, weighing 900 g, moving at 47 m/s. during
Ivanshal [37]

Solution :

Given :

Mass of the baseball, m = 200 g

Velocity of the baseball, u = -30 m/s

Mass of the baseball after struck by the bat, M = 900 g

Velocity of the baseball after struck by the bat, v = 47 m/s

According to the conservation of momentum,

Mv+mu=Mv_1+mv_2

(900 x 47) + (200 x -30)  = (900 x v_1) + (200 x v_2)

36300 =  (900 x v_1) + (200 x v_2)

9v_1 + 2v_2 = 363 ..............(i)

9v_1 = 363 - 2v_2

v_1=\frac{363 - 2v_2}{9}

The mathematical expression for the conservation of kinetic energy is

\frac{1}{2}Mv^2+\frac{1}{2}mu^2 = \frac{1}{2}Mv_1^2+\frac{1}{2}mv_2^2

\frac{1}{2}(900)(47)^2+\frac{1}{2}(200)(-30)^2 = \frac{1}{2}(900)v_1^2+\frac{1}{2}(200)v_2^2    ................(ii)

$(9)(14)^2+(2)(-30)^2 = (9)v_1^2+2v_2^2$  

21681 = 9v_1^2+2v_2^2

Substituting (i) in (ii)

21681= 9\left( \frac{363-2v_2}{9}\right)^2+2v_2^2

(363-2v_2)^2+18v_2^2=195129

(363)^2+18v_2^2-2(363)(2v_2)+(363)^2-195129=0

22v_2^2-145v_2-63360=0

Solving the equation, we get

v_2=96 \ m/s, -30 \ m/s

The negative velocity is neglected.

Therefore, substituting 96 m/s for v_2 in (i), we get

v_1=\frac{363-(2 \times 96)}{9}

     = 19

Thus, only impulse of importance is used to find final velocity.

8 0
3 years ago
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