The kinetic energy of an object with a mass of #2 kg# constantly changes from #8 J# to #136 J# over #4 s#. What is the impulse on the object at #1 s#?

1 Answer
Aug 20, 2016

Answer:

#vec J_(0 to 1) = 4(sqrt( 10) - sqrt( 2) )hat p# N s

Explanation:

I think there is something wrong in the formulation of this question.

With Impulse defined as
#vec J = int_(t = a)^b vec F(t) \ dt #
#= int_(t = a)^b vec dot p (t) \ dt = vec p(b) - vec p(a)#

then the Impulse on the object at t= 1 is

#vec J = int_(t = 1)^1 vec F(t) \ dt = vec p(1) - vec p(1) = 0#

It may be that you want the total impulse applied for #t in [0,1]# which is

#vec J = int_(t = 0)^1 vec F(t) \ dt = vec p(1) - vec p(0) qquad star#

To evaluate #star# that we note that if the rate of change of kinetic energy #T# is constant, ie:

#(dT)/(dt) = const #

then

#T= alpha t + beta#

#T(0) = 8 implies beta = 8#

#T(4) = 136 = alpha(4) + 8 implies alpha = 32#

#T= 32 t + 8#

Now #T = abs(vec p)^2/(2m)#.

#implies (vec p * vec p)= 4(32 t + 8)#

#vec p = 2sqrt( (32 t + 8)) hat p#

and
#vec p(1) - vec p (0)#

# = (2sqrt( (32 + 8)) - 2sqrt( 8) )hat p#

# = 4(sqrt( 10) - sqrt( 2) )hat p# N s