Sapling Learning Map m Before Collision nm 0.750v n,final Two particles approach

Question

Sapling Learning Map m Before Collision nm 0.750v n,final Two particles approach each other with equal and opposite speed v. The mass of one particle is m, and the of the other particle is n is just a unitless number. Snapshots of the system before, during, and after the elastic collision are shown above After the collision the first particle moves in the exact opposite direction with speed 0.750v, and the speed of the second particle is unknown. What is the value of n? Number ne Previous check Answer o Next d Exit Hint

Explanation / Answer

Apply conservation of momentum

m*v + n*m*(-v) = m*(-0.75)*v + n*m*vf

v - n*v = -0.75*v + n*vf

vf = (v - n*v + 0.75*v)/n

vf = (1.75 - n)*v/n ---------(1)

Apply conservation of kinetic energy

Kf = Ki

(1/2)*m*(0.75*v)^2 + (1/2)*n*m*vf^2 = (1/2)*m*v^2 + (1/2)*n*m*v^2

0.5625*v^2 + n*vf^2 = v^2 + n*v^2

n*vf^2 = v^2*(1 - 0.5625 + n)

n*vf^2 = v^2*(0.4375 + n)

vf = v*sqrt((0.4375 + n)/n) ------(2)

from equations 1 and 2

(1.75 - n)*v/n = v*sqrt((0.4375 + n)/n)

(1.75 - n)/n = sqrt((0.4375 + n)/n)

==> n = 0.778 <<<<<<<----------------Answer

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