In Problem 2, what is the speed of the car at (a) point A, (b) point B(c) point C?(d) How high will the car go on the last hill, which is too high for it to cross? (e) If we substitute a second car with twice the mass, what then are the answers to (a) through and (d)?

Short Answer

Expert verified

a) The speed of the car at pointA is17.0m/s

b) The speed of the car at pointB is26.5m/s

c) the speed of the car at pointC is33.4m/s

d) Final height56.7m

e) Even if a second car has twice the mass, it is evident that the above results do not depend on mass. Thus, a different mass for the coaster must lead to the same results.

Step by step solution

01

Given

i) Mass of carm=825​ kg

ii) The height h=42 m

iii) Gravitational acceleration g=9.8m/s2

iv) Initial velocityv0=17m/s

02

To understand the concept

The mechanical energy Emecof a system is the sum of its kinetic energy Kand potential energy U.

Formula:

K2+U2=K1+U1

K=12mv02

U=mgh

03

(a) Calculate the speed of the car at point A

At pointA,UA=mgh and at initial pointU0=mgh.

That means initial position of the car and point A is on same height, so,

UA=U0

K0+U0=KA+UA

K0=KA

∴role="math" localid="1661168162281" v0=vA=17.0m/s

04

(b) Calculate the speed of the car at point B

The height of pointB from bottom=h/2

UB=mgh2

From principle of conservation of mechanical energy,

K0+U0=KB+UB

12mv02+mgh=12mvB2+mgh2

We have to find VB, so rearrange the equation; we can get

vB2=2×(mgh2+12mv02)m

vB=2×(mgh2+12mv02)m

vB=(gh+v02)

vB=(9.80×42.0+17.02)

vB=700.6

vB=26.5m/s

05

(c) Calculate the speed of the car at point C

The height of pointfrom bottom=0

UC=0

From principle of conservation of mechanical energy,

K0+U0=KC+UC

12mv02+mgh=12mvC2+0

We have to find vB, so rearrange the equation; we can get

vC2=2×(mgh+12mv02)m

vC=2×(gh+12v02)

vC=(2gh+v02)

vC=(2×9.80×42.0+17.02)

vC=1112.2

vC=33.4m/s

06

(d) Calculate how high will the car go on the last hill, which is too high for it to cross

To find the “final” height, we setKf=0.In this case, we have

K0+U0=Kf+Uf

12mv02+mgh=12mvB2+mghf

hf=h+v022g

hf=42.0+1722×9.80

hf=56.7m

07

(e) Calculate the answers to (a) through and (d) if we substitute a second car with twice the mass

From the formula, we can see that the results derived above are independent of mass. So change in mass would not change the values found above.

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