1)An archer shoots an arrow toward a 300-g target that is sliding in her directi
ID: 1779799 • Letter: 1
Question
1)An archer shoots an arrow toward a 300-g target that is sliding in her direction at a speed of 2.00 m/s on a smooth, slippery surface. The 22.5-g arrow is shot with a speed of 38.5 m/s and passes through the target, which is stopped by the impact. What is the speed of the arrow after passing through the target?_____m/s
2)A railroad car of mass 2.70 104 kg moving at 3.05 m/s collides and couples with two coupled railroad cars, each of the same mass as the single car and moving in the same direction at 1.20 m/s.
(a) What is the speed of the three coupled cars after the collision? ______m/s
(b) How much kinetic energy is lost in the collision?________J
Explanation / Answer
Answer 1. From conservation of momentum :
Momentum P = m *v where m is mass and v is velocity. Initially the target had a momentum of 2.40 * 0.3 = 0.72 Ns It is now stationary so it altered its momentum by this amount.The arrow must have also altered its momentum by the same amount .The original momentum was 36 * 0.0225= 0.81 Ns. It has lost 0.72 so its new momentum is 0.09 Ns, as momentum is velocity * mass then velocity b= momentum / mass = 0.09 / 0.0225 = 4 m/s
Answer 2.
“couples” means all 3 cars are coupled together after the collision. So
Total mass after collision is = 3 * 3.3 * 10^4 kg = 99000
we know that Momentum is always conserved! so
Initial momentum is = 3.3 * 10^4 * 2.00 + 2 * 3.3 * 10^4 * 1.20 =145200
Final momentum is = 3 * 3.3 * 10^4 * v =99000 * v
for answer (a) What is the speed of the three coupled cars after the collision?
Final momentum = Initial momentum
3 * 3.3 * 10^4 * v = 3.3 * 10^4 * 2.00 + 2 * 3.3 * 10^4 * 1.20
Divide both sides by 3.3 * 10^4
3 * v = 2.00 + 2 * 1.20 = 4.4
v = (4.4 ÷ 3) m/s = 1.467 is the velocity of the 3 coupled cars after the collision!
for answer (b) How much kinetic energy is lost in the collision?
Initial KE = (½ * 3.3 * 10^4 * 2.00^2) + (½ * 2 * 3.3 * 10^4 * 1.20^2) = 113520
Final KE = ½ * 3 * 3.3 * 10^4 * (4.4/3)^2 =106480
so, Initial KE – Final KE = kinetic energy is lost in the collision
= 113520 - 106480 = 7040 J
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