How to find the total momentum before and after a collision?

Momentum And Satellite motion - year 12 physics?

• I have already calculated these questions. I need answers to them so please help me out so i can compare. Remember when doing these questions make it as simple for me to understand as to everything you have done step by step it would really help in case I get something wrong. Q1. My mass is 100kg. What is my weight on the earth's surface? Q2. Find the gravitational force of attraction between the earth and the sun given that the mass of the earth is 6.0x10^24, the mass of the sun is 1.99x10^30 and their mean distance apart is 1.496x10^11m. Q3. How high above the surface of the earth must we go before the value of the gravitational acceleration is reduced to 4.9ms^2. Q4. suppose that a satellite, mass m, is rotating in a fixed circular orbit of radius R around the earth, mass M. 1. Explain why the speed of the satellite is given by v=square root GM/R. 2. Derive an expression for the period T of the satellite in terms of G,m and R. 3. Calculate the height above the Earths surface of a satellite that has a period of 100 minutes. Q5. A truck of mass 1850kg travels west at 8m/sec and collides with a car of mass 750kg travelling south at 12m/sec. After the collision the car and the truck stick together. Assuming no external forces, calculate. 1. the momentum of the system before collision. 2. the momentum of the system after collision 3. The velocity of the car and truck after the collision. 4. The total initial kinetic energy and the total final kinetic energy.

• Answer:

  Q1. My mass is 100kg. my weight on the earth's surface = 100*9.81 = 981 N Q2. Find the gravitational force of attraction between the earth and the sun given that the mass of the earth is 6.0x10^24, the mass of the sun is 1.99x10^30 and their mean distance apart is 1.496x10^11m. F = (GMm)/d^2 = {(6.673*10^-11)*(6.0*10^24)*(1.99*10^30)… = 3.56*10^22 N Q3. How high above the surface of the earth must we go before the value of the gravitational acceleration is reduced to 4.9ms^2. GM/r^2 = 9.8 for GM/(r+h)^2 = 4.9 (r+h)/r = sq rt(2) or h = (√ 2 -1)*r = 0.414*6.0*10^24 m = 2.485*10^24 m or 2485*10^21 km Q4. suppose that a satellite, mass m, is rotating in a fixed circular orbit of radius R around the earth, mass M. 1. Explain why the speed of the satellite is given by v=square root GM/R. Centripetal force required for orbiting speed v, = mv^2/R = GMm/R^2 or v = square root (GM/R) 2. Derive an expression for the period T of the satellite in terms of G,m and R. v = (R*2*π)/T = sq rt[GM/R] or T = [2π/{sq rt(GM)}]*R^(3/2) 3. Calculate the height above the Earths surface of a satellite that has a period of 100 minutes. Radius, R of the orbit for satellite having 100 minutes time period is given by (6000)^2 = [4*(π^2)/(GM)]R^3or R^3 = [{(6.67&10^-11)(6.0*10^24)*36*(10^6)}/(4… = 365*10^18 or R = 7.15*10^6 m So height above earth's surface = (7.15 -6.37)*10^6 m = 7.765*10^5 m or 776.5 km Q5. A truck of mass 1850kg travels west at 8m/sec and collides with a car of mass 750kg travelling south at 12m/sec. After the collision the car and the truck stick together. Assuming no external forces, calculate. 1. the momentum, P of the system before collision. = sq rt[(1850*8)^2 + (750*12)^2] = 17322 N s 2. the momentum of the system after collision = 17322 Ns 3. The velocity of the car and truck after the collision = 17322/[1850+750] = 6.66 m/s 4. The total initial kinetic energy and the total final kinetic energy. Initial K E = 0.5*[1850*8^2 + 750*12^2] = 113200 J Final KE = 0.5*[1850+750]*6.66^2= 57662 J