🎉 Announcing Numerade's $26M Series A, led by IDG Capital!Read how Numerade will revolutionize STEM Learning

Like

Report

Numerade Educator

Like

Report

Problem 34 Easy Difficulty

Two skaters collide and grab on to each other on frictionless ice. One of them, of mass $70.0 \mathrm{kg},$ is moving to the right at 2.00 $\mathrm{m} / \mathrm{s}$ , while the other. of mass 65.0 $\mathrm{kg}$ , is moving to the left at 2.50 $\mathrm{m} / \mathrm{s}$ . What are the magnitude and direction of the velocity of these skaters just after they collide?

Answer

$v_{2}=0.167 \mathrm{m} / \mathrm{s}$ to the left

Discussion

You must be signed in to discuss.

Video Transcript

{'transcript': "well, we have two skaters. A and B skater is mass is 70 point kilogram. So oh, scatter is mass is equal. Dio 70.0 kilogram and skater bees masses 65 point cause, you know kilogram soon myself. The second scatter is 65.0 kilograms. Well taken right is the positive direction we have. A scatter is the lost he before the collision. Let's call this velocity. Be a one, right? It's before the collision. Then it's equal to to find his Igor Casino meter for a second and skater abusive. Lost it before the collision is we be one right? So we be one is equal to minus 2.50 meters per second, right? Well, we know the in collision off all kinds since there is no external net force on the system. The total momentum before the collision because the total momentum after the collision so be one that's initial momentum Musical. To be to our teacher is final momentum. Well, if the religion is between two particles A and B, then we have a one lists B B one. It's total initial momentum is equal to be a two less B B two. It's total final momentum. I came. Well, we know that momentum is mass. Times lost. E Let's call the secret in number one. Well, so we can ride the college in the region in the following form I m a times we day one are less MB times we be one right sequel to M A times we hate you read it too. Bliss MB and B times We'd be too We be too right. And let's call this in. Question number two All right, well, since the do scatters skate on a frictionless eyes then there is no external lead falls on the horizontal motion. Also after the collision secret grab onto each other so they have the same final lost e So we have ah b a to right. We had to is equal to we. B two is equal to we to right? No lips, uh, plugging numbers. Iniquity number two ah, 70 into Joe. This, uh, 65 65 times minus. To find five Syrian that is equal to ah, 70 times. We to unless ah, 65 times we do. And now solving for we to B two is equal to 70 multiplied by two. Let's 65 multiply by minus 2.50 divided by about 70 plus 65. Therefore, V two is equal to minus zero point 167 meters per second."}