
A 20kg body is pushed with 7N for 1.5 sec then with a force of 5N for 1.7 sec, and finally with a force of 10N for 3 sec, the total impulse applied to the body and change in velocity will be
A. \[49Ns,12.5m{s^{ - 1}}\]
B. \[49Ns,2.45m{s^{ - 1}}\]
C. \[98Ns,4.9m{s^{ - 1}}\]
D. \[4.9Ns,2.45m{s^{ - 1}}\]
Answer
584.7k+ views
Hint: In this question, initially, the body is at rest, and some forces are applied for a specific time interval; hence to find the total impulse, first find the impulse for each force applied and then add them up to find the total impulse on the body, where \[I = Ft\]
Complete step by step answer:
Mass of the body \[m = 20kg\]
\[
{F_1} = 7N \\
{t_1} = 1.5\sec \\
\]
\[
{F_2} = 5N \\
{t_2} = 1.7\sec \\
\]
\[
{F_3} = 10N \\
{t_3} = 3\sec \\
\]
As we know, the impulse is the force applied to an object for an interval of time given as\[I = Ft - - (i)\], hence the impulse on the body at
\[
{I_1} = {F_1}{t_1} \\
= 7 \times 1.5 \\
= 10.5Ns \\
\]
\[
{I_2} = {F_2}{t_2} \\
= 5 \times 1.7 \\
= 8.5Ns \\
\]
\[
{I_3} = {F_3}{t_3} \\
= 10 \times 3 \\
= 30Ns \\
\]
Now find the total impulse applied on the body, since
\[
{I_T} = {I_1} + {I_2} + {I_3} \\
= 10.5 + 8.5 + 30 \\
= 49Ns \\
\]
Hence the total impulse applied on the body is\[ = 49Ns\]
We already know that impulse is equal to change in momentum; hence we can say
\[I = m\left( {v - u} \right) - - (ii)\]
Now substitute the value of mass and the impulse in equation (ii) to find the change in velocity where the initial velocity of the body \[u = 0\]
\[
I = m\left( {v - u} \right) \\
49 = 20\left( {v - 0} \right) \\
\left( {v - 0} \right) = \dfrac{{49}}{{20}} \\
v = 2.45\dfrac{m}{{{s^2}}} \\
\]
Hence the change in velocity of the body will be \[ = 2.45\dfrac{m}{{{s^2}}}\]
Option (A) is correct
Note:Impulse is the change of momentum of an object when the object is acted upon by force for some interval of time. By using impulse, we can calculate the change in momentum, which is given by the formula \[I = m\vartriangle v\]. Students must note that when an impulse is applied to a body, it changes the momentum of the body and makes the body to move.
Complete step by step answer:
Mass of the body \[m = 20kg\]
\[
{F_1} = 7N \\
{t_1} = 1.5\sec \\
\]
\[
{F_2} = 5N \\
{t_2} = 1.7\sec \\
\]
\[
{F_3} = 10N \\
{t_3} = 3\sec \\
\]
As we know, the impulse is the force applied to an object for an interval of time given as\[I = Ft - - (i)\], hence the impulse on the body at
\[
{I_1} = {F_1}{t_1} \\
= 7 \times 1.5 \\
= 10.5Ns \\
\]
\[
{I_2} = {F_2}{t_2} \\
= 5 \times 1.7 \\
= 8.5Ns \\
\]
\[
{I_3} = {F_3}{t_3} \\
= 10 \times 3 \\
= 30Ns \\
\]
Now find the total impulse applied on the body, since
\[
{I_T} = {I_1} + {I_2} + {I_3} \\
= 10.5 + 8.5 + 30 \\
= 49Ns \\
\]
Hence the total impulse applied on the body is\[ = 49Ns\]
We already know that impulse is equal to change in momentum; hence we can say
\[I = m\left( {v - u} \right) - - (ii)\]
Now substitute the value of mass and the impulse in equation (ii) to find the change in velocity where the initial velocity of the body \[u = 0\]
\[
I = m\left( {v - u} \right) \\
49 = 20\left( {v - 0} \right) \\
\left( {v - 0} \right) = \dfrac{{49}}{{20}} \\
v = 2.45\dfrac{m}{{{s^2}}} \\
\]
Hence the change in velocity of the body will be \[ = 2.45\dfrac{m}{{{s^2}}}\]
Option (A) is correct
Note:Impulse is the change of momentum of an object when the object is acted upon by force for some interval of time. By using impulse, we can calculate the change in momentum, which is given by the formula \[I = m\vartriangle v\]. Students must note that when an impulse is applied to a body, it changes the momentum of the body and makes the body to move.
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