Find the speed of car when people jumping off from it

  • #1

Homework Statement


A stationary car with 4 people in it has mass 100 kg. The mass of each of the person is 75 kg. They jump off the car with speed 2 ms-1 relative to the car. Calculate the speed of the car if
a. 4 people jump at once
b. They jump one by one each time



Homework Equations


momentum and kinematics maybe


The Attempt at a Solution


1 = people
2 = car

a) m1u1+m2u2 = m1v1+m2v2
0 = 4 x 75 x 2 - 100 x v2
v2 = 6 ms-1

b) 1st person jumps
0 = 75 x 2 - 325 x v2
v2 = 6/13 ms-1

2nd person jumps
the speed of 2nd person relative to the car = 2 + 6/13 = 32/13 ms-1
m1 u1 + m2 u2 = m1 v1 + m2 v2
0 - 325 x 6/13 = 75 x 32/13 - 250 x v2
v2 = 87/65 ms-1

3rd person jumps
the speed of 3rd person relative to the car = 2 + 87/65 = 217/65 ms-1
m1 u1 + m2 u2 = m1 v1 + m2 v2
0 - 250 x 87/65 = 75 x 217/65 - 175 x v2
v2 = 117/35 ms-1

4th person jumps
the speed of 4th person relative to the car = 2 + 117/35 = 187/35 ms-1
m1 u1 + m2 u2 = m1 v1 + m2 v2
0 - 175 x 117/35 = 75 x 187/35 - 100 x v2
v2 = 69/7 ms-1

the final speed of the car is 69/7 ms-1

I am not sure whether my answers are correct. Can someone please help checking my answers? Thank you very much
 

Answers and Replies

  • #2
136
0
Part a is correct.

However, in part b, you have your final velocities (after the first person jumps) incorrect. Since, according to your equations, the car is pushed in the negative direction, the next person's v2 would be 2 m/s minus the car's velocity - not plus.
 
  • #3
My thinking was if the person moves in opposite direction to the car, the relative speed of the person to the car will be higher since the person and the car move away faster from each other. I don't understand why it should be minus, not plus. Can you explain it more to me? Thank you very much
 
  • #4
136
0
I understand why you may think that, but no. Keep in mind your inertial frame of reference - although I'm not sure that really explains it very well...

After the first person jumps (we'll say to the right, which we'll call positive), the car begins moving to the left - which means the car's velocity (a vector) is negative. Since the problem asks for speed (a scalar), though, it's a positive number. Therefore, when the next person jumps to the right at a speed of 2 m/s relative to the car - and the car is traveling in the negative direction - the person's initial velocity/speed is 2 m/s to the right minus whatever speed the car was traveling. Basically, the person (because he/she is on the car that is going left) is going left initially. Then they jump at 2 m/s to the right. And because the 2 m/s is relative, they don't actually travel the full 2 m/s to the right in an inertial (stationary) frame.

Think about it this way: if you're in a car and drop something straight down out of the window - 0 m/s horizontally - the object you drop is still going to travel in the same direction that you're traveling until it hits the ground. Why? Because it has velocity in your direction of travel. So, if you were to calculate its final position, you'd have to take into account that initial velocity.

It isn't until just after the person jumps that their relative velocities change. Were the person to jump in the same direction that the car was moving, then yes, you'd add the velocities together - but you'd end up with even different values. And while the problem isn't specific as to which direction(s) the people are jumping, it is implied that they all jump in the same direction - thus adding to the car's speed.
 
  • #5
I got it now

Thank you very much for the help
 

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