Answer :
Answer:
26i-28j
Explanation:
Mathematically, rate of change of something with respect to another is obtained through the operation called differentiation. Integration is the reverse process of differentiation.
We know that force is the rate of change of momentum.
that is,
F=[tex]\frac{dP}{dt}[/tex]
where,
P=momentum
t=time
dP=small change in momentum
dt=small change in time
dP=Fdt (multiplying both sides with dt)
To get change of momentum within a particular time interval, we have to integrate the above expression giving the necessary limits according to the problem.
Here, we have to calculate the change in momentum between 1.0s and 2.0s. So, the limit of integration is 1.0 to 2.0.
Thus,
[tex]\int\limits^P_p {} \, dP[/tex]=[tex]\int\limits^2_1 {F} \, dt[/tex]
where,
p=momentum at t=1s
P=momentum at t=2s
substituting F in the above equation,
[tex]\int\limits^P_p {} \, dP[/tex]=[tex]\int\limits^2_1 {26i-12jt^{2}} \, dt[/tex]
[tex]\int\limits^P_p {} \, dP[/tex] = P-p = Change in momentum between 1.0s and 2.0s
[tex]\int\limits^2_1 {26i-12jt^{2}} \, dt[/tex] = [tex]\int\limits^2_1 {26i} \, dt - \int\limits^2_1 {12t^{2}j } \, dt[/tex] (Since we can integrate each term of an expression separately)
[tex]\int\limits^2_1 {26i} \, dt - \int\limits^2_1 {12t^{2}j } \, dt[/tex] = [tex]26i(2-1) - 12j(\frac{2^{3} -1^{3} }{3} )[/tex]
=26i-28j
therefore the change in momentum between 1.0s and 2.0s is 26i-28j.
Note: the answer we got is a vector. Because momentum is a vector and therefore change in momentum is also a vector