speed speed speed speed VR aV #VL Mass m - 0.1 kg moves to the right with speed

Question

speed speed speed speed VR aV #VL Mass m - 0.1 kg moves to the right with speed v 0.32 m/s and collides with an equal mass initially at rest. After this inelastic collision the system retains a fraction 0.64 of its original kinetic energy. How much impulse (in units of N sec) does the mass originally at rest receive during the collision? Hints: All motion is in 1D. Ignore friction between the masses and the horizontal surface. You will probably need to use the quadratic formula to solve the resulting equations. Ve must be greater than Vi, since the masses can't pass through each other!

Explanation / Answer

here,

m = 0.1 kg

intial speed , v = 0.32 m/s

let the final speeds be vl and vr

using conservation of momentum

m * v = m * vl + m * vr

0.32 = vl + vr ...(1)

and

intial kinetic energy * 0.64 = final kinetic energy

(0.5 * m * v^2)0.64 = ( 0.5 * m * vl^2 + 0.5 * m * vr^2)

0.32^2 * 0.64 = vl^2 + vr^2 ....(2)

from (1) and (2)

vr = 0.24 m/s

vl = 0.075 m/s

the impulse originally at rest recives , I = m * ( vl - 0)

I = 0.1 * ( 0.24 - 0) kg.m/s

I = 0.024 kg.m/s

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