3. Three clay balls (m1=40 g, m2=50 g, m3=60 g) collide...

Name: 3. Three clay balls (m1=40 g, m2=50 g, m3=60 g) collide simultaneously and then stick together on a horizontal plane. Their velocities right before the collision are given in the figure. What is the velocity (magnitude and direction) of the resulting blob of clay right after the collision?
Uploaded: 2022-04-22T11:24:29-08:00
Duration: 9 min 25 s
Channel: Maitreya E
Description: 3. Three clay balls (m1=40 g, m2=50 g, m3=60 g) collide simultaneously and then stick together on a horizontal plane. Their velocities right before the collision are given in the figure. What is the velocity (magnitude and direction) of the resulting blob of clay right after the collision?

3. Three clay balls (m1=40 g, m2=50 g, m3=60 g) collide simultaneously and then stick together on a horizontal plane. Their velocities right before the collision are given in the figure. What is the velocity (magnitude and direction) of the resulting blob of clay right after the collision?

Transcript

0:00 Hello.

00:02 So it is given to us in the question that the masses of the clay bowls are m1 equals 40 grams, m2 equals 50 grams, and m3 equals 60 grams.

00:13 And their velocities make an angle, theta 1 equals 45 degree, theta 2 equals 60 degree, and theta 3 equals 90 degree with the x axis.

00:23 Now the velocities are v1 is equal to 3 meter per seconds, v2 equals 4 meter per second, and v3 equals.

00:30 Is 5 meter per second.

00:32 Okay.

00:33 Now we have to resolve these velocities into their x components as well as the y components or the horizontal components as well as the vertical components.

00:43 Okay.

00:44 So it can be resolved into x component by.

00:48 So the x component of v1, v1 x should be equal to v1 cosine theta 1 and substituting the values we get this is equal to 3 cosine 45 degrees and that is equal to 2 .12 meter per second.

01:14 Similarly, the x component of v2, v2x could be written to be equal to v2 cosine theta 2 and substituting the values, v2 and theta 2, we get this.

01:31 Should be equal to 2 meter per second and the x component of v3 that is v3 x should be equal to v3 cosine theta 3 okay now substituting the values here we get this to be equal to 0 okay similarly we can write the y components as well so the y component of v1 is v1 i equals v1 sine theta 1 and substituting the values we get this is equal to 3 sign 3 sine 45 degree and on calculating we get this to be equal to 2 .12 meter per second.

02:28 Okay similarly the y component of v2 would be equal to v2 sin theta 2 .2.

02:39 And substituting the values we get this to be equal to 3 .464 meter per second.

02:49 Okay, 3 .464 meter per second.

02:55 And the y component of v3, v3y could be written to be equal to v3 sine theta 3 and that should be equal to 5.

03:10 Meter per second.

03:14 Okay.

03:16 Now we'll be using the conservation of momentum here.

03:19 So the initial momentum along the x -axis should be equal to the initial, sorry, the final momentum along the y -axis according to the conservation of momentum.

03:34 Okay.

03:35 So we can write m1 plus m2 plus m3 multiplied by vfx which is the final velocity of the combined clay bowl should be equal to m1 v1 x plus m2 v2 x plus m2 v2 x plus m3 m3 x okay which is the momentum before collision okay now substituting the value we get 40 plus 50 plus 60 multiplied by vfx should be equal to 40 multiplied by 2 .12 plus 50 multiplied by 2 plus 0.

04:45 Okay...

Question

3. Three clay balls (m1=40 g, m2=50 g, m3=60 g) collide simultaneously and then stick together on a horizontal plane. Their velocities right before the collision are given in the figure. What is the velocity (magnitude and direction) of the resulting blob of clay right after the collision?

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