You and your friends are doing physics experiments on a frozen pond that serves
ID: 1911010 • Letter: Y
Question
You and your friends are doing physics experiments on a frozen pond that serves as a frictionless, horizontal surface. Sam, with mass 71.0kg , is given a push and slides eastward. Abigail, with mass 55.0 kg, is sent sliding northward. They collide, and after the collision Sam is moving at 31.0 degree north of east with a speed of 5.40m/s and Abigail is moving at 22.0 degree south of east with a speed of 9.40m/s . a) What was the speed of each person before the collision? Sam's speed: 11.4 m/s b)Abigail's speed: c)By how much did the total kinetic energy of the two people decrease during the collision?Explanation / Answer
Let M1 be mass of sam = 71.0 kg Let V1i be initial velocity of sam along east i.e. x axis Let M2 be mass of abigail = 56.0 kg Let V2i be initial velocity of abigail along north i.e. y axis Since on collision they do not stick do each other, this is an elastic collision Let V1f be final velocity of sam = 5.40 m/s at an angle of 37 degs from the x axis Let V2f be final velocity of abigail = 8.40 m/s at an angle of -21 degs from the x axis Since they are perpendicular, conservation of momentum is different. Therefore, we find the components Now, V1fx = 5.40* cos 37 V1fy = 5.40 * sin 37 V2fx = 8.40* cos 21 V2fy = 8.40*sin 21 Conserving momentum in x and y direction separately In the x direction only sam moves initially Hence, V1i * M1 = M1*V1fx + M2*V2fx Solve for V1i V1i = (M1*V1fx + M2* V2fx)/ M1 V1i = V1fx + M2*V2fx/M1 Substituting values we get: V1i = 5.40*cos37 + (56.0/71.0)*8.40*cos 21 V1i = 10.5 m/s along east Analyzing y motion: Only Abigail moves along y initially, V2i *M2= (M1*V1fy + M2*V2fy) solve for V2i V2i = (M1*V1fy + M2*V2fy)/M2 V2i = V2fy + (M1/M2)*V1fy Substituting values, V2i = 8.40*sin 21 +(71.0/56.0)*5.4*sin 37 V2i = 7.13 m/s north The kinetic energy (KE) is known for both before and after cases KE1i = 0.5* 71.0* 10.5^2 = 3913.875 J KE1f = 0.5*71.0*5.4^2 = 1035.18 J delta KE1 = 1035.18 - 3913.875 = -2878.695 J Hence, it decreases for Sam KE2i = 0.5* 56.0* (7.13)^2 KE2i = 1423. J KE2f = 0.5*56.0*8.4^2 KE2f = 1975.68 J delta KE2 = 1975.68 - 1423 = 552.68 J Hence, it increases for Abigail
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