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10. describe a real - life example where impulse is used to protect peo…

Question

  1. describe a real - life example where impulse is used to protect people.
  2. what happens to the momentum of a system if no external forces act on it?
  3. how does mass affect momentum if velocity stays constant?
  4. can an object have momentum if it is not moving? explain.
  5. what is the si unit of momentum?
  6. how do airbags in cars relate to impulse and force?
  7. a 4 kg object moves at 5 m/s. what is its momentum?
  8. a 2 kg ball is hit with a force of 10 n for 0.2 seconds. what is the impulse?
  9. a 3 kg cart moving at 6 m/s collides with a stationary 2 kg cart. after the collision, the 3 kg cart moves at 2 m/s. what is the velocity of the 2 kg cart after the collision? (assume elastic collision and conservation of momentum.)
  10. a baseball of mass 0.15 kg is pitched at 40 m/s and hit back at 50 m/s. what is the impulse delivered to the ball?
  11. a 5 kg object experiences an impulse of 20 ns. what is the change in velocity?

Explanation:

Step1: Define momentum formula

Momentum $p = mv$, where $m$ is mass and $v$ is velocity.

Step2: Solve question 16

Given $m = 4\ kg$ and $v=5\ m/s$. Substitute into formula: $p=4\times5 = 20\ kg\cdot m/s$.

Step3: Define impulse formula

Impulse $J = F\Delta t$, where $F$ is force and $\Delta t$ is time - interval.

Step4: Solve question 17

Given $F = 10\ N$ and $\Delta t=0.2\ s$. Substitute into formula: $J = 10\times0.2=2\ N\cdot s$.

Step5: Apply conservation of momentum for question 18

Initial momentum $p_i=m_1v_1=(3\ kg)\times(6\ m/s)=18\ kg\cdot m/s$.
Final momentum $p_f=m_1v_1'+m_2v_2'$, where $m_1 = 3\ kg$, $v_1' = 2\ m/s$, $m_2 = 2\ kg$.
By conservation of momentum $p_i=p_f$, so $18=3\times2 + 2v_2'$.
Solve for $v_2'$: $18 = 6+2v_2'$, then $2v_2'=18 - 6=12$, and $v_2' = 6\ m/s$.

Step6: Calculate impulse for question 19

Initial velocity $v_i=- 40\ m/s$ (pitched direction), final velocity $v_f = 50\ m/s$.
Impulse $J=\Delta p=m(v_f - v_i)$. Given $m = 0.15\ kg$.
$J=0.15\times(50-( - 40))=0.15\times90 = 13.5\ N\cdot s$.

Step7: Use impulse - momentum theorem for question 20

Impulse - momentum theorem $J=\Delta p=m\Delta v$.
Given $J = 20\ N\cdot s$ and $m = 5\ kg$.
Solve for $\Delta v$: $\Delta v=\frac{J}{m}=\frac{20}{5}=4\ m/s$.

Step8: Answer question 10

Airbags in cars are an example. When a car crashes, the airbag inflates. It increases the time of impact, reducing the force exerted on the passengers according to $F=\frac{J}{\Delta t}$ (since impulse $J$ is fixed in a collision).

Step9: Answer question 11

According to the law of conservation of momentum, if no external forces act on a system, the total momentum of the system remains constant.

Step10: Answer question 12

Since $p = mv$, when velocity $v$ is constant, momentum $p$ is directly proportional to mass $m$. So, as mass increases, momentum increases and vice - versa.

Step11: Answer question 13

No. Since $p = mv$, if $v = 0$, then $p=0$.

Step12: Answer question 14

The SI unit of momentum is kilogram - meter per second ($kg\cdot m/s$).

Step13: Answer question 15

Airbags increase the time $\Delta t$ of the collision between the passenger and the car's interior. According to $J = F\Delta t$, for a given impulse $J$ (determined by the change in momentum of the passenger), increasing $\Delta t$ reduces the force $F$ exerted on the passenger.

Answer:

  1. Airbags in cars increase impact time to reduce force on passengers.
  2. The total momentum of the system remains constant.
  3. Momentum is directly proportional to mass when velocity is constant.
  4. No, because $p = mv$ and if $v = 0$, $p = 0$.
  5. $kg\cdot m/s$
  6. Airbags increase impact time to reduce force for a given impulse.
  7. $20\ kg\cdot m/s$
  8. $2\ N\cdot s$
  9. $6\ m/s$
  10. $13.5\ N\cdot s$
  11. $4\ m/s$