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ch4aika [34]
3 years ago
13

A 25.0 kg bumper car moving to the right at 5.00 m/s overtakes and col-

Physics
1 answer:
kramer3 years ago
8 0

Explanation:

This problem bothers elastic collision.

Given data

Mass m1= 25kg

Initial velocity u1= 5m/s

Final velocity v1= 1.5m/s

Mass m2= 35kg

Initial velocity u2=?

Final velocity v2 = 4.5m/s

A. To find the initial velocity of the 35kg car, let us Apply the principle of conservation of energy

m1u1+m2u2= m1v1+m2v2

25*5+ 35*u2= 25*1.5+ 35*4.5

125+35u2= 37.5+157.5

125+35u2=195

35u2= 195-125

35u2= 70

u2= 2m/s

The initial velocity is 2m/s

B. Totally not kinetic energy before impact

KE= 1/2m1u1²+ 1/2m2u2²

KE= (25*5²)/2+ (35*2²)/2

KE= 625/2 +140/2

KE= 312.5+70

KE= 382.5J

Total kinetic energy after impact

KE=1/2m1v1²+ 1/2m2v2²

KE= (25*1.5²)/2 +(35*4.5²)/2

KE= 56.25/2 +708.75/2

KE=28.125 +354.375

KE= 382.5J

We can see that energy is conserved

Kinetic energy before and after impact remains unchanged

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If a rock is thrown upward on the planet mars with a velocity of 11 m/s, its height (in meters) after t seconds is given by h =
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(a) 3.56 m/s 
(b) 11 - 3.72a 
(c) t = 5.9 s 
(d) -11 m/s  
For most of these problems, you're being asked the velocity of the rock as a function of t, while you've been given the position as a function of t. So first calculate the first derivative of the position function using the power rule. 
y = 11t - 1.86t^2 
y' = 11 - 3.72t 
Now that you have the first derivative, it will give you the velocity as a function of t. 
(a) Velocity after 2 seconds. 
y' = 11 - 3.72t 
y' = 11 - 3.72*2 = 11 - 7.44 = 3.56 
So the velocity is 3.56 m/s  
(b) Velocity after a seconds. 
y' = 11 - 3.72t 
y' = 11 - 3.72a  
So the answer is 11 - 3.72a  
(c) Use the quadratic formula to find the zeros for the position function y = 11t-1.86t^2. Roots are t = 0 and t = 5.913978495. The t = 0 is for the moment the rock was thrown, so the answer is t = 5.9 seconds.  
(d) Plug in the value of t calculated for (c) into the velocity function, so: 
y' = 11 - 3.72a
 y' = 11 - 3.72*5.913978495
 y' = 11 - 22
 y' = -11 
 So the velocity is -11 m/s which makes sense since the total energy of the rock will remain constant, so it's coming down at the same speed as it was going up.
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2 years ago
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