
A force of 10 N acts on a body of mass 2 kg for 3s, initially at rest. Calculate:
(i) The velocity acquired by the body
(ii) Change in momentum of the body
Answer
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Hint:Calculate the acceleration produced in the body using Newton’s second law. Use the first kinematic equation to determine the final velocity of the body after 3 s. The momentum of the body is the product of its mass and its velocity.
Formula used:
Newton’s second law, \[F = ma\],
where, m is the mass and a is the acceleration.
First kinematic equation,\[v = u + at\],
where, u is the initial velocity, v is the final velocity and t is the time.
Momentum, \[p = mv\],
where v is the velocity.
Complete step by step answer:
We have given the force acting on the body \[F = 10\,{\text{N}}\] and the mass of the body is \[m = 2\,{\text{kg}}\].
(i) We can determine the acceleration of the body using Newton’s second law of motion as follows, \[a = \dfrac{F}{m}\]
Here, F is the force and m is the mass of the body.
Substituting \[F = 10\,{\text{N}}\] and \[m = 2\,{\text{kg}}\] in the above equation, we get,
\[a = \dfrac{{10}}{2}\]
\[ \Rightarrow a = 5\,{\text{m/}}{{\text{s}}^2}\]
Let’s determine the final velocity of the body using the first kinematic equation as follows,
\[v = u + at\]
Here, u is the initial velocity, v is the final velocity, a is the acceleration and t is the time.
Substituting \[u = 0\], \[a = 5\,{\text{m/}}{{\text{s}}^2}\] and \[t = 3\,{\text{s}}\] in the above equation, we get,
\[v = 0 + \left( 5 \right)\left( 3 \right)\]
\[ \therefore v = 15\,{\text{m/s}}\]
Therefore, the final velocity of the body is 15 m/s.
(ii) We can express the change in the momentum of the body as,
\[\Delta p = mv - mu\]
\[ \Rightarrow \Delta p = m\left( {v - u} \right)\]
Substituting \[m = 2\,{\text{kg}}\], \[v = 15\,{\text{m/s}}\] and \[u = 0\] in the above equation, we get,
\[\Delta p = \left( 2 \right)\left( {15 - 0} \right)\]
\[ \therefore \Delta p = 30\,{\text{kg}}\,{\text{m/s}}\]
Therefore, the change in the momentum of the body is 30 kg m/s.
Note:Students can determine the change in momentum of body using Newton’s second law of motion, \[F = \dfrac{{dp}}{{dt}}\]. Substituting 10 N for force and 3 s for time, we can get the change in momentum 30 kg m/s. We have taken the initial velocity as zero since the body was initially at rest.
Formula used:
Newton’s second law, \[F = ma\],
where, m is the mass and a is the acceleration.
First kinematic equation,\[v = u + at\],
where, u is the initial velocity, v is the final velocity and t is the time.
Momentum, \[p = mv\],
where v is the velocity.
Complete step by step answer:
We have given the force acting on the body \[F = 10\,{\text{N}}\] and the mass of the body is \[m = 2\,{\text{kg}}\].
(i) We can determine the acceleration of the body using Newton’s second law of motion as follows, \[a = \dfrac{F}{m}\]
Here, F is the force and m is the mass of the body.
Substituting \[F = 10\,{\text{N}}\] and \[m = 2\,{\text{kg}}\] in the above equation, we get,
\[a = \dfrac{{10}}{2}\]
\[ \Rightarrow a = 5\,{\text{m/}}{{\text{s}}^2}\]
Let’s determine the final velocity of the body using the first kinematic equation as follows,
\[v = u + at\]
Here, u is the initial velocity, v is the final velocity, a is the acceleration and t is the time.
Substituting \[u = 0\], \[a = 5\,{\text{m/}}{{\text{s}}^2}\] and \[t = 3\,{\text{s}}\] in the above equation, we get,
\[v = 0 + \left( 5 \right)\left( 3 \right)\]
\[ \therefore v = 15\,{\text{m/s}}\]
Therefore, the final velocity of the body is 15 m/s.
(ii) We can express the change in the momentum of the body as,
\[\Delta p = mv - mu\]
\[ \Rightarrow \Delta p = m\left( {v - u} \right)\]
Substituting \[m = 2\,{\text{kg}}\], \[v = 15\,{\text{m/s}}\] and \[u = 0\] in the above equation, we get,
\[\Delta p = \left( 2 \right)\left( {15 - 0} \right)\]
\[ \therefore \Delta p = 30\,{\text{kg}}\,{\text{m/s}}\]
Therefore, the change in the momentum of the body is 30 kg m/s.
Note:Students can determine the change in momentum of body using Newton’s second law of motion, \[F = \dfrac{{dp}}{{dt}}\]. Substituting 10 N for force and 3 s for time, we can get the change in momentum 30 kg m/s. We have taken the initial velocity as zero since the body was initially at rest.
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