Texts a 1500 kg car moving east at 17 ms collides with a...

Name: texts a 1500 kg car moving east at 17 ms collides with a 1800 kg car moving south at 15 ms and the two cars stick together take e to be the x axis a what is the total momentum magnitude and 93541
Uploaded: 2021-12-24T00:57:07-08:00
Duration: 11 min 12 s
Channel: Ivan Kochetkov
Description: texts a 1500 kg car moving east at 17 ms collides with a 1800 kg car moving south at 15 ms and the two cars stick together take e to be the x axis a what is the total momentum magnitude and 93541

Texts: A 1500-kg car moving east at 17 m/s collides with a 1800-kg car moving south at 15 m/s, and the two cars stick together. Take E to be the +x axis. a. What is the total momentum (magnitude and direction) of the two cars just before the collision? b. What is the momentum (magnitude and direction) of the two cars just after the collision? c. What is the velocity (magnitude and direction) of the two cars just after the collision? Show work, please.

Transcript

00:01 Cars are colliding the first car is moving to the east and the second car is moving to the south so when they collide they stick together and we have to calculate the velocity after the collision and change and loss relative loss in the kinetic energy let's do this so first let's sketch this collision so at the collision point yeah the first car had in momentum of m m1b1 or wish let's let's write it down as p1 and the second car had a momentum of p2.

00:59 They are directed as shown here and now we can introduce y and x axis and we can write down the conservation of the momentum so vector sum of p1 and p2 equals to p3 which is momentum after the collision.

01:22 In terms of here we just need to calculate the sum of these two vectors and this is a resulting vector which is the diagonal of this right triangle because p1 and p2 are perpendicular.

01:38 So p3 is shown here.

01:42 Let's say let's presume that this angle alpha is angle with is direction and we can calculate this alpha now.

02:05 Alpha equal, from this triangle, alpha equals, sorry, alpha equals to a tangent of p2 over p1.

02:25 Or it equals to a tangent of m2 v2 over m1 b1.

02:33 Let's calculate it numerically.

02:35 It equals to a tangent of m2 times v2 divided by m1 v1 so let's calculate this number it is 46 .6 degree so therefore it's 46 .6 degree to the south to the east direction in yeah basically this angle is to the the east direction in the east south quadrant.

04:20 Okay we found the angle alpha.

04:22 Now we have to calculate the velocity of the collision.

04:27 So this velocity u equals to p3 divided by m1 plus m2.

04:37 And as we can see from the triangle p3 is a vector sum of as p3 is a equals 2 square root of p1 squared plus p2 squared.

04:47 So therefore p3 is square root of m1 b1 squared plus m2 squared and it's all divided by the sum of m1 and m2.

05:00 Now let's calculate u.

05:02 It equals to in the denominator it is 1 ,500 plus 1 ,800.

05:15 And the nominator we have 1 ,500 times 17 squared plus 1800 times 15 squared and it's meters per second times kilogram.

05:40 Now let's calculate the result...