Question 1 2 pts A distant, stationary meteor mysteriously...

Question 1 2 pts A distant, stationary meteor mysteriously explodes into two pieces. Piece A has a larger mass than Piece B. The two pieces of the meteor move apart from one another. Consider the meteor (and its subsequent two pieces) to be the system, and ignore any gravitational forces due to all bodies outside the system. Which of the following statements are true? (Check all that apply.) The objects must have equal speeds Mechanical Energy is conserved in this explosion The objects must have equal but opposite momenta Object A's speed is larger than object B's speed The objects must move in opposite directions

Transcript

00:01 This problem, i'm going to use energy conservation that is minus g m -e -m divided by r -e -plus r plus half m v square is equals to minus g -m -e -m divided by re plus half m v -dash square.

00:24 So this v -dash we would be finding.

00:26 Now cancel m m on lhs and rhs.

00:31 Take out minus gme common, that would be 1 by re plus r minus 1 by re plus half v square is equals to half v dash square.

00:46 Now by taking the lcm, we will be getting g mer divided by re into re plus r plus r half v square is equal to half v -dash square, multiply by two on both sides.

01:04 So, v -dash -square is equals to 2 gm -e are divided by re -e into re -e plus r plus v.

01:17 This is the v -dash -square value.

01:20 Squaring on both sides gives v -dash is equals to under root 2 -g -m -e -r, divided by r .e.

01:31 Into r .e.

01:33 Plus r.

01:34 Plus v.

01:35 Sv.

01:35 Now substitute the values of g, m .e, r, r, r e, and v.

01:40 That is 2 into 6 .67 into 10 .5 .97 into 10 power minus 11 into 5 .97 into 10 power 24 into 21 ,300, divided by 6378 into 6378 plus 21 ,300 into 10 ,000 into 10 power 3 plus 2 ,000 square.

02:07 This is completely under root.

02:10 Now, after solving this route, we would be getting v -dash is equals to 10 kilometers per second.

02:16 This is the value of the velocity.

02:20 Question we are asked to find out the impulse that is delta p now delta p is equals to so first understand situation one is at six kilometers per second in this direction and other is at two kilometers per second in this direction now to get the relative speed we need to add the speeds so delta b is equals to mass 109 into 10 power 6 into 6 plus 2 into 10 power 3 that gives delta p is equals to 72 into 10 power 9 kg meters per second.

02:58 So this is the value of impulse in this question.

03:03 The third question we are asked to find out the force required.

03:07 So force required is delta p that is impulse divided by time taken...

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