A marble of mass m is dropped through a fluid, starting from rest, and experienc

Question

A marble of mass m is dropped through a fluid, starting from rest, and experiences an upward resistive force of R = - bv in addition to a downward force due to gravity mg. Using the equation for the speed of the marble with respect to time, v(t) = mg/b (1 - e^-bt/m) find an expression for the force on the particle as a function of time. Treat time t = 0 as the moment the marble is released from rest. Using this expression for the force, find an expression for the impulse over the interval of time from t = 0 to t = T. Use the definition of momentum in terms of mass and velocity and the given function for the marble's speed v(t) to confirm your expression for impulse in the previous part.

Explanation / Answer

part a:

acceleration=a=dv/dt

==>a=(m*g/b)*(b/m)*exp(-b*t/m)

==>a=g*exp(-b*t/m)

hence force =mass*acceleration=m*g*exp(-b*t/m)

part b:

impulse=integration of F.dt

=integration of m*g*exp(-b*t/m)*dt

=m*g*exp(-b*t/m)/(-b/m)

=(-m^2*g/b)*exp(-b*t/m)

using the limits for t from t=0 to t=T:

impulse applied=(m^2*g/b)*(1-exp(-b*T/m))

part c:

impulse=change in momentum

change in momentum =mass*(final velocity-initial velocity)

=mass*(velocity at T - velocity at 0)

=m*(m*g/b)*((1-exp(-b*T/m))-(1-1))

=(m^2*g/b)*(1-exp(-b*T/m))

hence change in momentum is same as impulse applied.

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