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grin007 [14]
2 years ago
9

A 1050 kg sports car is moving westbound at 13.0 m/s on a level road when it collides with a 6320 kg truck driving east on the s

ame road at 12.0 m/s. The two vehicles remain locked together after the collision. What is the velocity (magnitude) of the two vehicles just after the collision?
At what speed should the truck have been moving so that it and car are both stopped in the collision?
Find the change in kinetic energy of the system of two vehicles for the situations of part A.
Find the change in kinetic energy of the system of two vehicles for the situations of part C.
Physics
1 answer:
Mila [183]2 years ago
6 0

Answer:

a) v_{3} =8.43 m/s

b) v_{2}=2.15m/s

c) ΔK=-28.18x10^4J

d)ΔK=-10.33x10^4J

Explanation:

From the exercise we know that there is a collision of a sports car and a truck.

So, the sport car is going to be our object number 1 and the truck object number 2.

m_{1}=1050kg\\v_{1}=-13m/s\\m_{2}=6320kg\\v_{2}=12m/s

Since the two vehicles remain locked together after the collision the final mass is:

m_{3}=7370kg

a) To find the velocity of the two vehicles just after the collision we must use linear's momentum principle

p_{1}=p_{2}

m_{1}v_{1}+ m_{2}v_{2}=m_{3}v_{3}

v_{3}=\frac{m_{1}v_{1}+m_{2}v_{2}}{m_{3}}=\frac{(1050kg)(-13m/s)+(6320kg)(12m/s)}{7370kg}

v_{3}=8.43m/s

b) To find the speed the truck should have had so both vehicles stopped in the collision we need to use the same principle used before

m_{1}v_{1}+ m_{2}v_{2}=0

v_{2}=\frac{-m_{1}v_{1}}{m_{2} }=\frac{-(1050kg)(-13m/s)}{(6320kg)}=2.15m/s

c) To find the change in kinetic energy we need to do the following steps:

ΔK=k_{2}-k_{1}=\frac{1}{2}m_{3}v_{3}^{2}-(\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2} )

ΔK=\frac{1}{2}(7370)(8.43)^{2}-(\frac{1}{2}(1050)(-13)^{2}+\frac{1}{2}(6320)(12)^{2} )=-28.18x10^{4}J

d) The change in kinetic energy where the two vehicles stopped in the collision is:

ΔK=k_{2}-k_{1}=0-(\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2} )

ΔK=-(\frac{1}{2}(1050)(-13)^{2}+\frac{1}{2}(6320)(2.15)^{2} )=-10.33x10^4J

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Given that in a parallel circuit:

R1 = 12 ohms

R2= 15 ohms

I = 12 A

I2 = 4 A

V=?

R=?

R3 =?

P=?

Since,

V= IR

or,

V2 = I2 * R2

V2= 4* 15

V2 = 60V

Since in a parallel circuit voltage remain same in all component of the circuit and is equal to the source voltage.

Therefore,

V= V1 = V2 = V3 = 60V

Since,

V= IR

R= V/I

R= 60/12

R= 5 ohm

That is total resistance is equal to 5 ohms.

Since for parallel circuit,

1/R= 1/R1 + 1/R2 + 1/R3

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or

1/R3= 1/5- 1/12- 1/15

1/R3= 1/20

or

R3= 20 ohms

Since,

V=IR

I= V/R

I1= V1/ R1

I1= 60/12

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I3= V3/R3

I3= 60/20

I3= 3A

Since,

P=VI

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P1= 60* 5

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P2= 60* 4

P2= 240watt

P3= V3*I3

P3= 60*3

P3= 180 watt

Hence we have,

R1= 12 ohms , R2= 15 ohms, R3= 20 ohms, R= 5 ohms

I1= 5A, I2= 4A, I3= 3A, I= 12 A

V1= V2= V3= V= 60V

P1= 300 watt, P2= 240 watt, P3 = 180 watt, P= 720 watt



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