Momentum, Collisions and Explosions Legacy Problem #32 Guided Solution

An effective problem solver has the habit of reading the problem carefully and developing a mental picture of what's going on. A diagram is a great way to do that. An effective problem solver identifies known and unknown information, maybe even listing it on the diagram, and then uses physics understanding in order to plot out a strategy to get from known to unknown information. Here in this problem I have what I call a two-dimensional collision. The two objects involved in the collision are not moving in the same direction, nor in the opposite direction, but rather are moving at right angles to one another. The fullback has a mass of 92 kg and is a moving solid with a speed of 5.8 m per second. I represent the fullback by an arrow that is directed down on a page of paper. I label the arrow fullback and I list m and v values. m equals 92 kg and v equals 5.8 m per second. The fullback collides with a lineman who is running west, and I represent the lineman by an arrow that is heading off to the left on my page of paper. I label it lineman and I give it an m and a v value. m equals 110 kg and v equals 3.6 m per second. Now my unknown in this question is the speed of the two players after the collision. It's a tackle, so we can assume that they collide, they hang on to each other, and they travel in the same direction as though a single object after the collision. To find this value of the post-collision speed and direction of the two players, I'm going to have to use momentum conservation. I'm going to have to find the total or the sum of the momentum of the two players before the collision and say it's equal to the sum of the two players after the collision. Now when I'm summing vectors, I always have to take into account their direction and use principles of vector addition. Here for the fullback moving south and the halfback or the lineman moving to the west, what I'll have to do is add them up using the Pythagorean Theorem. So here we go, employing our strategy. We're going to first determine the momentum of the lineman. A mass of 110 kg and a velocity of 3.6 m per second would give the lineman a momentum of 396 kg m per second to the west. For the fullback, the southerly momentum can be calculated by taking the 92 kg mass, multiplying by the 5.8 m per second velocity, we'd get 533.6 kg m per second. Since these aren't right angles to one another, adding them together to find the total system momentum before the collision involves the use of the Pythagorean Theorem. So I'll say 396 squared plus 533.6 squared is equal to the total momentum squared, and I'll solve for the total momentum. When I do that, I get 664.4885 kg m per second. Now to determine the velocity after the collision, what I have to presume is that momentum is going to be conserved, so the momentum I just calculated for before the collision is the same as the momentum after the collision. And after the collision, this total momentum is shared by both objects moving in the same direction. If we determine the total mass of the two objects, we would get 202 kg. That's 110 plus 92 kg. So what I'm going to do is take the total momentum and divide by the total mass. Now I'm using the equation P equals mv, and into the equation I'm putting total values. And as I do, that's going to give me the velocity of these two players together after the collision. They're moving with a velocity of 3.2895 m per second. I'll round that to two significant digits. The answer is 3.3 m per second. Now I have to find the direction of the two players after the collision. You might recall, we have previously calculated their momentums individually, and then we determine the total by summing them up as vectors. Returning to the diagram in which we added these two at right angles to one another, I'm going to draw a triangle, a momentum triangle, in which I draw an arrow to the left. That's from the linemen, and I label it 396 kg m per second. Then I draw an arrow down, starting from the arrowhead of my first arrow, and I label it 533.6 kg m per second. Now I draw a resultant, a sum of these two vectors. It's the hypotenuse of a right triangle that has as its sides these two individual momentum values. And what I want to do is find the direction of this hypotenuse. So the angle that the hypotenuse makes with West is what I'm going to call theta. And I'm going to calculate the angle theta. Doing so means I'll have to use the tangent function, and I'll say the tangent of theta is equal to the side opposite, 533.6, divided by the side adjacent, 396. When I do that and solve for theta, I get 53.41 degrees. Rounded, it's 53 degrees. That's a vector that's 53 degrees rotated south of the original West direction. So I describe the direction as 53 degrees south of West.