Physics inelastic collision problem?
A 56kg ice skater traveling at 4.0m/s to the north suddenly grabs the hand of a 65kg skater traveling at 12.0m/s in the opposite direction as they pass. The two skater continue skating together with joined hands. A. what is the final velocity of the two skaters? i got 4.6 m/s to the south for an answer i know part A is right. B. What is the decrease in kinetic energy during the collision? i cant get part B. can some please explain 'how' to do it. i already know the answer from the back of my textbook but i cant figure out how to get it. ive tried like 3 different ways. please help!
Inelastic/elastic? Collision problem?
Because the ball bounced and you are not holding it, we must assume elastic collision. That means that both Momentum and Kinetic Energy is conserved. So, we need only calculate original KE and divide that between the you and the ball. We know the KE of the ball before and after, so we give you what ever is left. KEbb = 0.5MV^2 = 0.5 * 0.4kg*(10 m/s)^2 = 20 kg-m/s^2 (Joule) KEBa = 0.5*0.4*(8 m/s)^2 = 12.8 kg-m/s^2 (Joule) So your KE must be (20 - 12) = 8 kg-m/s^2 KEy =0.5 * 70kg*V^2 8 kg-m/s^2 = 0.5 * 70 V^2 V^2 = 0.228 (m/s)^2 V= 0.47 m/s = 47 cm/s QED
2D collisions problem?
The masses are not provided because they are identical and can simply be represented as M in our conservation of momentum equations: 1. horizontally: M*21cm/s - M*46cm/s = M*-13cm/s + M*V2_xf → mass M cancels V2_xf = -12 cm/s ◄ 2. vertically M*35cm/s + M*15cm/s = M*10cm/s + M*V2_yf V2_yf = 40 cm/s ◄ 3. initial KE = Σ ½mv² = ½M*(21² + 35² + 46² + 15²)cm²/s² = 2003.5M cm²/s² final KE = ½M(13² + 10² + 12² + 40²)cm²/s² = 1006.5M cm²/s² < CORRECTIONS to both KE values> % KE conserved = final / initial * 100% = 50.2% so the % lost = 49.8% ◄ corrected UPDATE: #4 and 5 look like #1 and #2 to me. Uncertainty is not in my wheelhouse, but I'd say ± 0.5 cm/s Hope this helps!
Help, Physics Ice Skater Problem?
Two ice skaters, Daniel (mass 69.5 kg) and Rebecca (mass 49.9 kg) are practicing. Daniel stops to tie his shoelace and, while at rest, is struck by Rebecca, who is moving at 13.0 m/s before she collides with him. After the collision, Rebecca has a velocity of magnitude 6.4 m/s at an angle of 52.5° from her initial direction. Both skaters move on the frictionless, horizontal surface of the rink. (1) What are the magnitude and direction of Daniel's velocity after the collision? __m/s (2) ___° (clockwise from Rebecca's original direction of motion) (3) What is the change in total kinetic energy of the two skaters as a result of the collision? All I need is the answer, thanks!
Conservation of Momentum Problem?
A 0.350-kg ice puck, moving east with a speed of 5.54 m/s , has a head-on collision with a 0.950-kg puck initially at rest. Assume that the collision is perfectly elastic. A) What is the speed of the 0.350-kg puck after the collision? B) What is the direction of the velocity of the 0.350-kg puck after the collision? C) What is the speed of the 0.950-kg puck after the collision? D) What is the direction of the velocity of the 0.950-kg puck after the collision?
I'll pick best answer for help on this one physics problem. Please help?
Along the original line of motion : Total momentum before collision = Total momentum after collision 0.5 × 2 = 0.5 × v₁ × cos30° + 0.5 × v₂ × cos60° v₁ × [(√3)/2] + v₂ × (1/2) = 1 (√3)v₁ + v₂ = 2 ...... [1] Along the direction perpendicular to the original line of motion : Total momentum before collision = Total momentum after collision 0 = 0.5 × v₁ × sin30° + 0.5 × v₂ × sin60° v₁ × (1/2) + v₂ × [(√3)/2] = 0 v₁ = (√3)v₂ ...... [2] Put [2] into [1] : (√3)(√3)v₂ + v₂ = 2 4v₂ = 2 v₂ = 0.5 m/s Put v₂ = 0.5 into [2] : v₁ = (√3) × 0.5 v₁ = (√3)/2 = 0.866 m/s Total kinetic energy before collision = (1/2) × 0.5 × 2² = 1 J Total kinetic energy after collision = (1/2) × 0.5 × (0.5)² + (1/2) × 0.5 × (0.866)² = 0.25 J Total kinetic energy after collision < Total kinetic energy before collision Thus, the collision was NOT elastic.
You find yourself stranded out on this impossibly slick ice. Collision physics problem.?
You find yourself stranded out on this impossibly slick ice. There is no friction so you can't walk at all. You have a 2.5 kg physics book so you throw it away from yourself and it has a speed of 8.4 m/s. how much time does it take for YOU to reach the other side of the ice which is 15.5 m away? Your mass is 42 kg and gravity is 10 m/s.
What types of problems can ships and boats have while they are sailing?
Collision (striking a moving object) or allision (striking a stationary object).Striking marine debris (flotsam, jetsam, deadhead logs, etc)Grounding on shoals, rocks or shore.Loss of propulsive power or steering control.Fire or explosion.Taking water through hull (crack, hole, through-hull connections) or by downflooding (waves, immersion of deck edge), leading to foundering, capsize or sinking.Capsize from wave action, from wind or from shifting weights aboard.Failure of electrical power (affects navigation equipment, communications eqt, propulsion, etc.)Failure of hydraulic power (affects steering, windlass, cargo eqt., etc.)Piracy and other types of criminal activity.Acts of war.Mutiny. and other forms of crewing problems.Disorientation due to reduced visibility (fog, rain, snow, sea-smoke)Injuries and illness to crew, or personnel going overboard.Cargo shifting aboard (or overboard) or other damage due to cargo (e.g. rice expands when wet and can split open the hull).Structural damage from waves, from metal fatique, from corrosion or rot.Out of fuel, or becalmed due to lack of wind.Stranded in ice, damage from ice.Financial problems of owners.The most common problem: human error in all its forms.
A 63.0-kg ice hockey goalie (Physics)?
The total momentum of the system is a conserved quantity. Equating the total momentum before and after the collision for objects with mass mₐ and mᵦ mₐ vᵢₐ + mᵦ vᵢᵦ = mₐ vfₐ + mᵦ vfᵦ ----------> [i and f denoting initial and final velocities] or initial momentum = final momentum This equation is valid for any 1-dimensional collision. Note that, assuming we know the masses of the colliding objects, the above equation only fully describes the collision given the initial velocities of both objects, and the final velocity of at least one of the objects. An elastic collision is one in which the total kinetic energy of the two colliding objects is the same before and after the collision. For an elastic collision, kinetic energy is conserved. That is: 0.5 mₐ vᵢₐ² + 0.5 mᵦ vᵢᵦ² = 0.5 mₐ vfₐ² + 0.5 mᵦ vfᵦ² The collision is fully specified given the two initial velocities and masses of the colliding objects. Combining the above equations gives a solution to the final velocities for an elastic collision of two objects: vfₐ = [(mₐ - mᵦ) vᵢₐ + 2 mᵦ vᵢᵦ]/[mₐ + mᵦ] vfᵦ = [2 mₐ vᵢₐ − (mₐ - mᵦ) vᵢᵦ]/[mₐ + mᵦ] substituting the values mₐ = 63 kg mᵦ = 0.140 kg vᵢₐ = 0 m/s vᵢᵦ = 41 m/s => vfₐ = [(63 - 0.140) 0 + 2(0.140)41]/[63.140] = 0.181818 m/s ------------------->(in the original direction of the puck) and => vfᵦ = [0 − (63 - 0.140)41]/[63.140] = -40.81818 m/s (opposite direction) answers can be confirmed with the below online calculator hope this helps
Could we terraform Mars by redirecting comets into collisions?
Sure, the problem is the shear amount of cometary material needed is mind blowing. To give Mars oceans worth of water would require at least 1*10^17 tons of water. Crashing 1 billion tons of water on to mars, every day for 273,972 years would acheive that.Lets assume we need nitrogen to provide enough inert gas in the terraformed Mars’s atmosphere. Lets say 10 kPa worth, That would be equal to 6.5*10^14 tons of nitogen (based on multiplications of the present mass of mars’s atmosphere divided by molecular weight diffrence), or 8*10^14 tons of ammonia. Crashing 1 billion tons of ammonia, every day for 2178 years would acheive that.But wait that is assuming comets of pure water or pure ammonia, in reality comets might at best be 10% that! But the raw energy needed to move a comet from the keplar belt to Mars would be less then the energy to extract water and ammonia and ship them in a pure form. Better yet if they are going to be extracted and purified, might as well get it from the much closer asteroid belt instead. Ceres for example woud have far more than enough water, and the delta-V to get from Ceres to Mars is ~1 km/s, verse the >3 km/s needed to go from somewhere beyond Pluto to Mars.