Momentum and impulse Worked Solutions — Further Maths Maths

Fully worked, step-by-step solutions to Further Maths Momentum and impulse questions. See exactly how to solve problems on impulse-momentum-principle, impulse, sign-convention, impulse-of-a-constant-force.

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Further Maths70 questionsStep-by-step solutions
Question 1
2 markseasy
A particle is moving in a straight line on a smooth horizontal plane and is modelled as a particle. Take the direction of motion of the particle immediately before the impulse acts as the positive direction. The mass of the particle is m=0.5 kgm=0.5\ \text{kg}. Its velocity immediately before it receives an impulse is u=8 m s−1u=8\ \text{m s}^{-1} and its velocity immediately afterwards is v=−2 m s−1v=-2\ \text{m s}^{-1}. Find II, the impulse received by the particle, in N s\text{N s}.

Worked solution

  1. State the impulse-momentum principle

    I=mv−mu=m(v−u)I=mv-mu=m\left(v-u\right)

    The impulse acting on the particle equals its change in momentum.

  2. Substitute the given values

    I=0.5×(−2)−0.5×8I=0.5\times\left(-2\right)-0.5\times8

    The mass is multiplied by each velocity, keeping the signs.

  3. State the final answer with its units

    I=−5 N sI=-5\ \text{N s}

    This is the impulse, quoted with its units.

Answer
I=−5 N sI=-5\ \text{N s}
Question 2
2 markseasy
A particle is moving in a straight line on a smooth horizontal plane and is modelled as a particle. Take the direction of motion of the particle immediately before the impulse acts as the positive direction. The mass of the particle is m=2 kgm=2\ \text{kg}. Its velocity immediately before it receives an impulse is u=3 m s−1u=3\ \text{m s}^{-1} and its velocity immediately afterwards is v=7 m s−1v=7\ \text{m s}^{-1}. Find II, the impulse received by the particle, in N s\text{N s}.

Worked solution

  1. State the impulse-momentum principle

    I=mv−mu=m(v−u)I=mv-mu=m\left(v-u\right)

    The impulse acting on the particle equals its change in momentum.

  2. Substitute the given values

    I=2×7−2×3I=2\times7-2\times3

    The mass is multiplied by each velocity, keeping the signs.

  3. Evaluate the change in momentum

    I=14−6=8I=14-6=8

    The final momentum minus the initial momentum gives the impulse.

  4. State the final answer with its units

    I=8 N sI=8\ \text{N s}

    This is the impulse, quoted with its units.

Answer
I=8 N sI=8\ \text{N s}
Question 3
2 markseasy
A particle is moving in a straight line on a smooth horizontal plane and is modelled as a particle. Take the direction of motion of the particle immediately before the impulse acts as the positive direction. The mass of the particle is m=0.4 kgm=0.4\ \text{kg}. Its velocity immediately before it receives an impulse is u=15 m s−1u=15\ \text{m s}^{-1} and its velocity immediately afterwards is v=−5 m s−1v=-5\ \text{m s}^{-1}. Find II, the impulse received by the particle, in N s\text{N s}.

Worked solution

  1. State the impulse-momentum principle

    I=mv−mu=m(v−u)I=mv-mu=m\left(v-u\right)

    The impulse acting on the particle equals its change in momentum.

  2. Substitute the given values

    I=0.4×(−5)−0.4×15I=0.4\times\left(-5\right)-0.4\times15

    The mass is multiplied by each velocity, keeping the signs.

  3. State the final answer with its units

    I=−8 N sI=-8\ \text{N s}

    This is the impulse, quoted with its units.

Answer
I=−8 N sI=-8\ \text{N s}
Question 4
2 markseasy
A particle is moving in a straight line on a smooth horizontal plane and is modelled as a particle. Take the direction of motion of the particle immediately before the impulse acts as the positive direction. The mass of the particle is m=3 kgm=3\ \text{kg}. Its velocity immediately before it receives an impulse is u=2 m s−1u=2\ \text{m s}^{-1} and its velocity immediately afterwards is v=−4 m s−1v=-4\ \text{m s}^{-1}. Find II, the impulse received by the particle, in N s\text{N s}.

Worked solution

  1. State the impulse-momentum principle

    I=mv−mu=m(v−u)I=mv-mu=m\left(v-u\right)

    The impulse acting on the particle equals its change in momentum.

  2. Substitute the given values

    I=3×(−4)−3×2I=3\times\left(-4\right)-3\times2

    The mass is multiplied by each velocity, keeping the signs.

  3. State the final answer with its units

    I=−18 N sI=-18\ \text{N s}

    This is the impulse, quoted with its units.

Answer
I=−18 N sI=-18\ \text{N s}
Question 5
2 markseasy
A particle of mass m=4 kgm=4\ \text{kg} is moving in a straight line. Take the direction of motion of the particle at the initial instant as the positive direction. Its velocity at that instant is u=3 m s−1u=3\ \text{m s}^{-1}. A constant force F=6 NF=6\ \text{N} acts on the particle for a time t=2 st=2\ \text{s}. The particle is modelled as a particle and the plane is smooth. Find vv, the velocity of the particle at the end of this time, in m s−1\text{m s}^{-1}.

Worked solution

  1. State the impulse-momentum principle

    I=mv−mu=m(v−u)I=mv-mu=m\left(v-u\right)

    The impulse acting on the particle equals its change in momentum.

  2. State the impulse of a constant force acting for a time tt

    I=FtI=Ft

    The impulse of a constant force is the force multiplied by the time for which it acts.

  3. Find the impulse delivered by the force

    I=Ft=6×2=12I=Ft=6\times2=12

    The force is constant, so the impulse is simply FtFt.

  4. State the final answer with its units

    v=6 m s−1v=6\ \text{m s}^{-1}

    This is the velocity of the particle after the force has acted.

Answer
v=6 m s−1v=6\ \text{m s}^{-1}

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