On a frictionless horizontal air table, puck A (with mass 0.253 kg ) is moving toward puck B (with mass 0.374 kg ), which is initially at rest. After the collision, puck A has velocity 0.125 m/s to the left, and puck B has velocity 0.650 m/s to the right.
Part A: What was the speed vAi of puck A before the collision?
Part B:Calculate ΔK, the change in the total kinetic energy of the system that occurs during the collision.
Collisions in one dimension?
2018-01-09 11:22 am
回答 (3)
2018-01-09 11:50 am
Take rightward velocity to be positive, leftward velocity to be negative.
A.
Before collision: m₁ = 0.253 kg, u₁ = vAi (m/s), m₂ = 0.374 kg, u₂ = 0 m/s
After collision: m₁ = 0.253 kg, v₁ = -0.125 m/s, m₂ = 0.374 kg, v₂ = 0.650 m/s
Conservation of momentum: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
0.253u₁ + 0.374*0 = 0.253*(-0.125) + 0.374*0.650
Speed of puck A before the collision, uAi = u₁ = 0.836 m/s
B.
The change of total kinetic energy of the system, ΔK
= m₁v₁ + m₂v₂ - m₁u₁ - m₂u₂
= (1/2)m₁v₁² + (1/2)m₂v₂² - (1/2)m₁u₁² - (1/2)m₂u₂²
= (1/2)*0.253*(-0.125)² + (1/2)*0.374*0.650² - (1/2)*0.253*0.836² - (1/2)*0.374*0² J
= -0.00743 J
A.
Before collision: m₁ = 0.253 kg, u₁ = vAi (m/s), m₂ = 0.374 kg, u₂ = 0 m/s
After collision: m₁ = 0.253 kg, v₁ = -0.125 m/s, m₂ = 0.374 kg, v₂ = 0.650 m/s
Conservation of momentum: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
0.253u₁ + 0.374*0 = 0.253*(-0.125) + 0.374*0.650
Speed of puck A before the collision, uAi = u₁ = 0.836 m/s
B.
The change of total kinetic energy of the system, ΔK
= m₁v₁ + m₂v₂ - m₁u₁ - m₂u₂
= (1/2)m₁v₁² + (1/2)m₂v₂² - (1/2)m₁u₁² - (1/2)m₂u₂²
= (1/2)*0.253*(-0.125)² + (1/2)*0.374*0.650² - (1/2)*0.253*0.836² - (1/2)*0.374*0² J
= -0.00743 J
2018-01-09 12:00 pm
Consider velocity towards right as positive
Conservation of momentum
mA*vAi + mB*vBi = mA*vAf + mB*vBf
0.253*vAi + 0.374*0 = 0.253*(-0.125) + 0.374*(0.650)
0.253*vAi = 0.253*(-0.125) + 0.374*(0.650)
vAi = [0.253*(-0.125) + 0.374*(0.650)] / 0.253
vAi = 0.836 m/s to right
ΔK = Final KE - Initial KE
ΔK = (½mA*vAf² + ½mB*vBf²) - (½mA*vAi² + ½mB*vBi²)
ΔK = (½0.253*(-0.125)² + ½0.374*0.650²) - (½0.253*0.836² + ½0.374*0²)
ΔK = -0.007426 J
There was a loss of 7.426 mJ
Conservation of momentum
mA*vAi + mB*vBi = mA*vAf + mB*vBf
0.253*vAi + 0.374*0 = 0.253*(-0.125) + 0.374*(0.650)
0.253*vAi = 0.253*(-0.125) + 0.374*(0.650)
vAi = [0.253*(-0.125) + 0.374*(0.650)] / 0.253
vAi = 0.836 m/s to right
ΔK = Final KE - Initial KE
ΔK = (½mA*vAf² + ½mB*vBf²) - (½mA*vAi² + ½mB*vBi²)
ΔK = (½0.253*(-0.125)² + ½0.374*0.650²) - (½0.253*0.836² + ½0.374*0²)
ΔK = -0.007426 J
There was a loss of 7.426 mJ
2018-01-09 11:58 am
A. ( MV)ai +(MV)bi = ( MV)af +(MV)bf
(0.253)Vai + 0 = (0.253)(-0.125) + (0.374)(+0.650) ... solve for Vai
Vai = [(0.253)(-0.125) + (0.374)(+0.650)] / 0.253 = 0.836 m/s to the right
B. K energy = (1/2)mv^2 <<<< squaring v makes all KE positive, so you can ignore signs for direction in this calculation
delta K = (1/2)(0.253)(0.125^2) + (1/2)(0.374)(0.650^2) - (1/2)(0.253)(0.836^2)
.... use a calculator
(0.253)Vai + 0 = (0.253)(-0.125) + (0.374)(+0.650) ... solve for Vai
Vai = [(0.253)(-0.125) + (0.374)(+0.650)] / 0.253 = 0.836 m/s to the right
B. K energy = (1/2)mv^2 <<<< squaring v makes all KE positive, so you can ignore signs for direction in this calculation
delta K = (1/2)(0.253)(0.125^2) + (1/2)(0.374)(0.650^2) - (1/2)(0.253)(0.836^2)
.... use a calculator
收錄日期: 2021-04-18 18:01:01
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20180109032246AAogj2q