An experiment is performed in deep space with two uniform...

An experiment is performed in deep space with two uniform spheres, one with mass 30.0 kg and the other with mass 103.0 kg . They have equal radii, r = 0.10 mm . The spheres are released from rest with their centers a distance 45.0 mm apart. They accelerate toward each other because of their mutual gravitational attraction. You can ignore all gravitational forces other than that between the two spheres. Part A When their centers are a distance 25.0 m apart, find the speed of the 30.0 kg sphere. Part B Find the speed of the sphere with mass 103.0 kg. Part C Find the magnitude of the relative velocity with which one sphere is approaching to the other. Part D How far from the initial position of the center of the 30.0 kg sphere do the surfaces of the two spheres collide? Express your answer using two significant figures.

Transcript

00:01 Hello everyone it is given in the question that mass of first sphere is 30kg and mass of second sphere is 103k and it is given that radius of both the spheres are same and that is 0 .10 mm right and it is given that initial separation between the two spaces is here 45m so right here that initial separation this is 45 amm now finally it is 25 meter.

00:35 So basically we have to find the speed of 30kg plus where speed of 103kg, relative velocity and final position.

00:46 So coming to the solution front.

00:55 Since we all know that here we'll apply the m1b1 plus m2b2 is equal to 0 and second indication is half l1 b1 square plus half m2 v2 is square minus of g m m upon r this is equal to minus g lm 2 upon r this is equal to minus g lm 2 upon l now here l is equal to 45 m so this is equal to 10 h d minus t t t t t t t t t t t t t multiplied so for equation 1 and we name it as equation number 2 and this is a equation number 1.

01:43 So we can say that v2 is equal to minus m1 v1 upon m2.

01:50 Now we'll put this value in the equation number 2 and finally we'll get here this as b1 square is equal to 2 times of g m2, 1 a 1 a 1 a x minus 1 of 1 upon 1 plus m1 upon so we can write here 2 gm 2 upon and that is here l upon x minus 1 upon m 2 now we'll put the values so we're going to be equal to root over 2 times of g so g here this is 6 .67 multiplied by 10 to 4 minus 11 that is again multiplied by 10 to 4 3 basically mass is here when we'll write 1 1 that is nm 25 that is further multiplied by l upon x minus 1 upon x minus 1 and am 2 is here 1003 now this we were coming out to be 189 point 17 multiplied by 10 0 .11 so this is equal to 1 .307 multiplied by 10 x2 minus 5 meter per second now here speed of 30 kg space speed on 30 kages is equal to 1 .3 is not required by 10x2 per minus 5 meter 2nd.

04:12 Now coming to the b part that is speed of the second speed...

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