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#?

Question

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#?

Sebastian Acosta · Accepted Answer

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

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

Layla Ahmed · Answer

undefined To find the impulse at 1 second, use the impulse-momentum theorem: Impulse = Change in momentum. Since impulse equals the change in kinetic energy, Impulse = Final kinetic energy at 1 s - Initial kinetic energy. Initial kinetic energy is 8 J, and final kinetic energy at 1 s is 64 J. Thus, Impulse = 64 J - 8 J = 56 J. Therefore, the impulse on the object at 1 second is 56 J.