View Full Version : Momentum problem
psycovic23
Dec20-04, 11:02 AM
"An atomic nucleus initially moving at 420 m/s emits an alpha particle in the direction of its velocity, and the new nucleus slows to 350 m/s. If the alpha particle has a mass of 4.0u and the original nucleus has a mass of 222 u, what speed does the alpha particle have when it is emitted?"
I did 222u * 420 m/s + 0 = 222u * 350m/s + 4 u * X m/s and I get 3885 m/s, but the answer says it's 4200. What did I do wrong?
assyrian_77
Dec20-04, 11:19 AM
The mistake is in the RHS of your equation. The new nucleus does not have mass 222 u anymore, its new mass is 218 u.
psycovic23
Dec20-04, 11:23 AM
Ah, I see. Thanks!
psycovic23
Dec20-04, 12:23 PM
Here's another one....
A 1.0x10^3 kg Toyota collides into the rear end of a 2.2x10^3 kg Cadillac stopped at a red light. THe bumpers lock, the brakes are locked, and the cars skid forward 2.8m before stopping. The coefficient of kinetic friction between the tires are .40. What's the speed of the Toyota at impact?
I know that the equation is
1.0 * 10^3 v_1 = (1.0x10^3 + 2.2 * 10^3) v'
and that the force lost to friction is
3.2*10^3 kg * g * .40 = 35,123N
but I don't know how to find the v` of the equation...as I don't know where to get a second equation.
v_1 - v_2 = v_2' - v_1'
doesn't work since it just ends up equaling 0...how do I do this?
\vec{F}_{friction}=m_{total}\vec{a}
Any equations of motion you can use that include displacement, acceleration, and velocities?
psycovic23
Dec20-04, 12:57 PM
Ah.......I get it! Tricky :tongue2: Thank you!
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