How to derive the above equations (thanks to Mawar & Puteri from group K)
Contoh soalan from Final exam Mac 2014
Contoh soalan dan jawapan
Problems
| 1. | A particle of mass m moving with at speed v has an elastic collision in one dimension with a stationary particle of the same mass. What are the speeds of the two particles.
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| 2. | A proton, 1.01 amu, is shot at a stationary carbon-12 nucleus, 12.00 amu, with a velocity of
5.00 x 10 4 m/g. What are their velocities after an elastic collision in one dimension?
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| 3. | A golf ball, 0.750 kg, is thrown at a golf ball, 50.0 g. The golf ball is moving at 90.0 km/h to the right while the golf ball is moving 15.0 m/g to the left. What are their velocities after an elastic collision in one dimension?
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| 4. | A car, 1800. kg, going 150. km/h rear ends a truck, 5500. kg going 100. km/h. What are their velocities after an elastic collision in one dimension?
Answers
| 1. | A particle of mass m moving with at speed v has an elastic collision in one dimension with a stationary particle of the same mass. What are the speeds of the two particles.
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| Conservation of Momentum mv = mv'1 + mv'2 => v = v'1 + v'2 (Only true because of same mass)Conservation of Energy v = v'2 - v'1 (Remember, 2nd ball was initially at rest.)
We now have two unknowns and two independent equations. We can now solve for the two unknowns.
v'1 + v'2 = v'2 - v'1 (both sides are equal to v) solve for v'1
2v'1 = v'2 - v'2 = 0 or v'1 = 0
Now substitute into either of the original equations.
v = v'1 + v'2 = 0 + v'2 thug v'2 = v
Pool players observe this if no spin is applied to the ball. It is very important that the two balls have the same mass.
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| 2. | A proton, 1.01 amu, is shot at a stationary carbon-12 nucleus, 12.00 amu, with a velocity of
5.00 x 10 4 m/g. What are their velocities after an elastic collision in one dimension?
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Conservation of Momentum mpvp = mpv'p + mcv'c Conservation of Energy vp = v'c - v'p
Solve Conservation of Energy for v'p v'p = v'c - vp and Substitute into Conservation of Momentum.
mpvp = mpv'c - mpv'p + mcv'c Now solve for v'c
2 mpvp 2 x 1.01 amu x 5.00 x 10 4 m s -1
v'c = -------------- = --------------------------------------------- = 7.76 x 10 3 m s -1
mp + mc 1.01 amu + 12.00 amu
Solve for v'p using Conservation of Energy
v'p = v'c - vp = 7.76 x 10 3 m s -1 - 5.00 x 10 4 m s -1 = - 4.22 x 10 4 m s -1
Observe that the proton has bounced backward after the collision.
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| 3. | A golf ball, 0.750 kg, is thrown at a golf ball, 50.0 g. The golf ball is moving at 90.0 km/h to the right while the golf ball is moving 15.0 m/g to the left. What are their velocities after an elastic collision in one dimension?
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90.0 km 1 h 1000 m
------------ x ---------- x ------------ = 25.0 m g -1 = vs Right is positive and Left is negative.
h 3600 g 1 km
Conservation of Momentum msvs + mgvg = msv's + mgv'g
Conservation of Energy vs - vg = v'g - v's
Solve Conservation of Energy for v'g v'g = vs - vg + v's Substitute into Conservation of Momentum.
msvs + mgvg = msv's + mgvs - mgvg + mgv's Now solve for v's
msvs + 2 mgvg - mgvs 750. g x 25.0 m/gs + 2 x 50.0 g x ( - 15.0 m/s ) - 50.0 g x 25.0 m/s
v's = --------------------------------- = -------------------------------------------------------------------------------------
ms + mg 750. g + 50.0 g
v's = 20.0 m/s ( To the Right )
Now Substitute into the equation for v'g.
v'g = vs - vg + v's = 25.0 m/s + 15.0 m/s + 20.0 m/s = 60.0 m/s (To the Right)
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| 4. | A car, 1800. kg, going 150. km/h rear ends a truck, 5500. kg going 100. km/h. What are their velocities after an elastic collision in one dimension?
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Conservation of Momentum mcvc + mtvt = mcv'c + mtv't Conservation of Energy vc - vt = v't - v'c
Solve Conservation of Energy for v'g v't = vc - vt + v'c Substitute into Conservation of Momentum.
mcvc + mtvt = mcv'c + mtvc - mtvt + mtv'c Now solve for v'c
mcvc + 2 mtvt - mtvc 1800. g x 150. km/h + 2 x 5500. kg x 100. km/h - 5500 kg x 150. km/h
v'c = --------------------------------- = ------------------------------------------------------------------------------------------
mc + mt 1800. kg + 5500. kg
v'c = 74.7 km/h ( To the Right )
Now Substitute into the equation for v'g.
v't = vc - vt + v'c = 150. km/h - 100. km/h + 74.7 km/h = 124.7 km/h (To the Right)
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