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A 2190 kg car moving east at 10.7 m/s collides with a 3210 kg car moving east. The cars stick together and move east as a unit after the collision at a velocity of 5.46 m/s. a) What is the decrease in kinetic energy during the collision

Sagot :

45152 Joules

Explanation

In physics, the kinetic energy of an object is the energy that it possesses due to its motion.its formula is

[tex]\begin{gathered} E_k=\frac{1}{2}mv^2 \\ \text{where m is the mass of the object and v its sp}eed \end{gathered}[/tex]

also, A collision in which the objects stick together is sometimes called perfectly inelastic because it reduces internal kinetic energy more than does any other type of inelastic collision,

When two objects collide under inelastic colliition, the final velocity with which the object moves is given by

[tex]\begin{gathered} V=\frac{(M_1V_1+M_2V_2}{M_1+M_2} \\ \end{gathered}[/tex]

Where,

V= Final velocity

M1= mass of the first object in kgs

M2= mas of the second object in kgs

V1= initial velocity of the first object in m/s

V2= initial velocity of the second object in m/s

then, replace

[tex]\begin{gathered} V=\frac{(M_1V_1+M_2V_2}{M_1+M_2} \\ 5.46=\frac{(2190\cdot10.7)+(3210\cdot V_2)}{2190+3210} \\ 5.46=\frac{23433+(3210\cdot V_2)}{5400} \\ 5400\cdot5.46=23433+(3210\cdot V_2) \\ 29484=23433+(3210\cdot V_2) \\ 29484-23433=3210V_2 \\ 6051=3210V_2 \\ \frac{6051}{3210}=\frac{3210V_2}{3210} \\ 1.88=V_2 \end{gathered}[/tex]

it means the velocity of the second car is 1.88 m/s

Step 2

a)Now, let's find the total kinetic energy before the collition

[tex]\begin{gathered} E_k=\frac{1}{2}mv^2 \\ E_1=\frac{1}{2}(2190kg)(10.7)^2 \\ E_1=1252366\text{ Joules} \\ \text{now, the car 2} \\ E_2=\frac{1}{2}(3210)(1.88)^2 \\ E_2=5672 \end{gathered}[/tex]

so, the total is

[tex]\begin{gathered} E_{total\text{ }}\text{= 125366 Joules +5672.712 JOules} \\ E_{total\text{ }}\text{=131038} \end{gathered}[/tex]

b) now, the total energy after the collition

Let

m=m1+m2=2190+3210=5400kg

v=5.46

replace

[tex]\begin{gathered} E_k=\frac{1}{2}mv^2 \\ E_t=\frac{1}{2}(5400)(5.46)^2 \\ E_t=85885.92\text{ Joules} \end{gathered}[/tex]

C) finally, the decrease of kinetic energy is the difference to those values

[tex]\text{Decrease}=131038-85556=45152.08\text{ Joules}[/tex]

hence, the answer is 45152 Joules

I hope this helps you

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