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Almost all of you might have played with Angry birds. Isnt it fun? Lets try to learn some physics using this game. There are other interesting articles explaining the physics concepts using Angry birds. The Physics of Angry Birds. Introducing projectile motion using Angry Birds. Angry Birds and Physics. Angry Birds in the Physics Classroom.
Here I have created set of questions that will help you explain physics concepts like newton's laws of motion, motion in a straight line, projectile motion etc using Angry Birds. Hopefully, next time when you see a question with projectile motion in your exams (IIT JEE, AIEEE or board exam) you will remember this and solve the question :) Lets assume that the sizes of bird and pig are negligible. Q1. Simple question to start with. Let the mass of the bird be m. The distance of the pig to targeted is d. It is given that on full stretch [by what we call gulel in hindi) the velocity of the bird is v0. What should be the angle of projection to hit the pig? Here I have drawn figure to explain (i am really bad with drawing).
Q2. What the the maximum height achieved by the bird in this flight? Q3. What will be the final velocity of the the angry bird hitting the pig? Q4. Lets say, in second level, the pig stands on an elevation as shown in the figure. Now, what should be the angle of projection of the bird, maximum height achieved by the bird and final velocity of the bird when it hits the pig? Given that pig is sitting at heigh h. Remember that there will be two scenarios - bird can hit when it is ascending or when it is descending.
Q5. Same question as Q4, except that now the angry bird starts from an elevation (of height h) and the pig is on the ground. How will the trajectory be now? In next set of questions - I will bring in conservation of momentum (when the angry bird splits into three parts) and external force coming in when angry bird is in the flight. |
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Let me try to attempt this - v0 can be split into two components [ remember v0 is a vector and vector can be analyzed independently across coordinates ]
We know from the conservation of energy principle that energy is always conserved. Lets apply this in vertical direction. You can easily figure out that at the max height of the object, the vertical velocity will be zero. When the object is thrown with velocity v0sin(theta), the gravitational force acts in opposite direction and so the velocity decreases. Ultimately it becomes zero and then starts becoming negative, i.e., it starts coming down. Also note that there is no acceleration in horizontal direction, so no change in energy (starting kinetic energy == final kinetic energy) Knowing this, lets write the energy (vertical component) when the object is thrown = potential energy + kinetic energy = [m x g x 0 ] + [(1/2) x m x (v0sin theta) ^2 ] (potential energy is zero) = (1/2)m(v0sin theta)^2 Energy at the top (maximum height achieved) = [m x g x h] + [(1/2) x m x 0] (kinetic energy is zero in vertical direction) = mgh By law of conservation of energy, mgh = (1/2)m(v0sin theta)^2 => gh = (1/2)(v0sin theta)^2 (equation 1) Let us also find the time taken by this object to reach at the maximum height. v(final) = 0 = v0sin(theta) - gt => t = v0sin(theta)/g (equation 2) Solving the equation in x direction s = ut + (1/2)aT^2 [fundamental equation of motion] Please note that there is no acceleration working in the horizontal direction, so we can write d = v0 x cos(theta) x T Also note that the time of going to maximum height will be same as time of coming down (downward acceleration is same) So we can say T = 2t, we get d = v0 x cos(theta) x 2t (equation 3) Replace the value of t from equation 2 in equation 3. d = v0 x cos(theta) x 2v0sin(theta)/g => sin(theta)cos(theta) = dg/2v0^2 => sin(2theta)/2 = dg/2v0^2 [ use maths equation sin(2a) = 2 sina cosa => theta = (1/2) sin^-1 (dg/v0^2) d is given, v0 is given and g is known. This leaves only theta as variable. This solves question 1 Plug this in equation 1 and you get the maximum height attained. This solves question 2. The third question is easy and can be solved without solving any equation. The horizontal velocity is constant. The vertical velocity goes down because of g acting downwards and goes to zero and then when it starts falling down it goes negative. Since the height gained [initial velocity to zero] is same as height lost [zero velocity to final]. The final vertical component of vertical velocity will be exactly same as the initial component but in the opposite direction. So the final velocity in vector notion = v(final) = vx - vy = v0cos(theta) - v0sin(theta). However, the magnitude of this velocity is same as v0. Q4 and Q5 after the break :) perhaps it is easier to explain if I give some values to d, v0 etc.
(Jan 09 '12 at 18:43)
answerer
Lets put some value to get a feel of the answer derived. Lets d = 10 m, assume g = 10ms^2 for simplicity and v0 = 10 meters per sec. Time of flight = T = 2t (as we discussed in the previous answer) = 2 v0sin(theta)/g = 2x10sin(theta)/10 = 2 sin(theta) Horizontal distance covered d = v0 x cos(theta) x T = 10cos(theta)x2 sin(theta) => 10 = 10x2 cos(theta)sin(theta) => 1 = sin(2theta) => 2theta = 90 degrees => theta = 45 degrees Find h from the equation derived from conservation of energy gh = (1/2)(v0sin theta)^2 h = (1/2x10) [10/srqt(2)]^2 = 2.5 meters
(Jan 10 '12 at 10:08)
answerer
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Lets solve Q4 now, when the pig is at elevation, there are two ways the bird can hit the pig, while going up or while coming down. Lets assume that it hits while coming down. Lets write down the equation of conservation of energy (assuming max height be H, and final velocity is vf) energy at point of origin = energy at max height = energy at the final destination (pig) mg(0) + (1/2)m[v0sin(theta)]^2 = mgH + 0 = mgh + (1/2)mVfy^2 [ Vfy is y component of Vf] => (1/2)[v0sin(theta)]^2 = gH = gh + (1/2)Vfy^2 Total time of flight will be time to reach the max height (t1) + time to reach from max height to height h (t2) v(final) = 0 = v0sin(theta) - gt1 => t1 = v0sin(theta)/g H- h = 0 + (1/2) g t2^2 => t2 = sqrt[2(H-h)/g] Total time of flight = t1 + t2 Solve equation in horizontal direction (no acceleration, remember?) d = v0 x cos(theta) x (t1+t2) => d = v0 cos(theta) x [v0sin(theta)/g + sqrt(2(H-h)/g)] You know d, v0, H, h and g, solve this equation to get the value of theta. Once we know that we can easily find the value of final velocity from law of conservation of energy, the max height we have already found out. Remember, we assumed that the bird strikes the pig while descending, but what about when it strikes while ascending? Of course we can write all the equation again for this case, but can we use the symmetry to get the answer? |
Now with Angry Birds Space, we will get even more Physics concepts to learn :) Looking forward to explanations.