1 to 2 Interactions, or explosions

Click For Summary
SUMMARY

The discussion revolves around a physics problem involving conservation of momentum. The scenario describes Tim, with a mass of 50.00 kg, jumping off a skateboard weighing 2.00 kg, initially traveling at 1.70 m/s. To solve for Tim's jump velocity, participants emphasize using the momentum equation, specifically that momentum (p) equals mass (m) times velocity (v). The key takeaway is that the total momentum before Tim jumps must equal the total momentum after he jumps, with the skateboard's final velocity being zero.

PREREQUISITES
  • Understanding of conservation of momentum
  • Familiarity with the momentum equation (p = mv)
  • Basic knowledge of mass and velocity concepts
  • Ability to differentiate between weight and mass
NEXT STEPS
  • Study detailed examples of conservation of momentum problems
  • Learn how to apply the momentum equation in various scenarios
  • Explore the differences between mass and weight in physics
  • Investigate real-world applications of momentum in sports and engineering
USEFUL FOR

Students studying physics, educators teaching momentum concepts, and anyone interested in applying conservation laws in practical scenarios.

lucky47
Messages
7
Reaction score
0
1 to 2 Interactions, or "explosions"

I'm stuck doing this problem on my physics homework.. it's different than the type of problems we have been doing in class, and he hasn't taught us the equation to solve this question. The problem is: Tim, mass 50.00 kg, is riding a skateboard, mass 2.00 kg, traveling at 1.70 m/s. Tim jumps off and the skateboard stops dead in its tracks. With what velocity did he jump?

Can anyone tell me what equation to use in order to get the answer? :confused:
 
Physics news on Phys.org
Think of momentum...

The skateboard has a cerain momentum of 'mv'. What velocity would Tim have to jump off of the skateboard to cause the final velocity of the board to be zero?

Use conservation of momentum before and after the jump (remember Tim has initial momentum).

This is neglecting the vertical component of Tim's jump.

Regards,
Sam
 
Hm.. I'm not quite getting it. So to figure out the momentum of Tim, it is 50kg, his weight; times the velocity of the skateboard, 1.7m/s, correct?
And now the momentum of the skateboard is 2kg times 1.7m/s..
Now to find the velocity i take the two numbers I got from solving the above and divide it by the mass of the skateboard?

Am I remotely correct in what I am doing?
 
Firstly, let's clear something up... Momentum is the product of velocity and MASS (not weight: weight is a force).

You have worked out the total momentum before the jump, but note that the velocity of the skateboard is now zero, so dividing by its mass is not useful.

Tim after the jump however, has a momentum mv (you know his mass), what is v?

Sam.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
905
  • · Replies 38 ·
2
Replies
38
Views
5K
  • · Replies 6 ·
Replies
6
Views
5K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
3
Views
2K
  • · Replies 12 ·
Replies
12
Views
4K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
4K
Replies
5
Views
1K