Physics, 26.11.2019 04:31 silveryflight

The force on a particle of mass m is given by f=26i-12t2j where f is in n and in s. what
will be the change in the particle’s momentum between 1.0s and 2.0s

Answers: 1

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Answers

Answer from: sarahhN7534

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= $\frac{dP}{dt}$

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,

$\int\limits^P_p {} \, dP$ = $\int\limits^2_1 {F} \, dt$

where,

p=momentum at t=1s

P=momentum at t=2s

substituting F in the above equation,

$\int\limits^P_p {} \, dP$ = $\int\limits^2_1 {26i-12jt^{2}} \, dt$

$\int\limits^P_p {} \, dP$ = P-p = Change in momentum between 1.0s and 2.0s

$\int\limits^2_1 {26i-12jt^{2}} \, dt$ = $\int\limits^2_1 {26i} \, dt - \int\limits^2_1 {12t^{2}j } \, dt$ (Since we can integrate each term of an expression separately)

$\int\limits^2_1 {26i} \, dt - \int\limits^2_1 {12t^{2}j } \, dt$ = $26i(2-1) - 12j(\frac{2^{3} -1^{3} }{3} )$

=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

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The force on a particle of mass m is given by f=26i-12t2j where f is in n and in s. what
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