# An object of mass 1 kg is traveling with a constant speed of 10 m/s in the positive x-direction. Consider case 1 in which the object above comes to a stop in 1 second. Consider case 2 in which the same object comes to a stop in 10 seconds instead of 1 second.

Which case had the bigger change in momentum?

Which case had the bigger impulse?

I know impulse is J = f*(Tf-Ti) but I don't see any forces listed. The time is bigger for one but maybe the force was less? They could be the same, or it could be "not enough info"?

## in both cases, the object has the same initial and final momentum

check the definition of impulse

## Well, let me juggle this question for you. In both cases, the initial velocity is the same, but the time it takes for the object to come to a stop is different. Since we're talking about a constant force here, we can use the equation F = m*a, where m is the mass and a is the acceleration.

In case 1, if the object comes to a stop in 1 second, the acceleration required would be much larger compared to case 2, where it takes 10 seconds for the object to come to a stop. So, in case 1, the force would be higher while in case 2, the force would be lower.

Now, when it comes to momentum, we know that momentum is equal to mass times velocity (p = m*v). Since the mass remains constant in both cases and the final velocity is zero because the object comes to a stop, we can conclude that the change in momentum will be the same in both cases.

However, when it comes to impulse, which is the product of force and time, we have a different story. In case 1, the force is higher but the time is shorter, whereas in case 2, the force is lower but the time is longer. So, to determine which case has the bigger impulse, we would need more information about the forces involved in both cases.

Therefore, we have reached the grand conclusion that there is "not enough info" to determine which case had the bigger impulse. Now, let's just hope this answer doesn't leave you in a state of confusion-fusion!

## To determine which case had the bigger change in momentum, we need to calculate the final momentum and subtract it from the initial momentum for both cases.

Case 1:

Initial momentum (p1) = mass (m) * velocity (v) = 1 kg * 10 m/s = 10 kg m/s

Final momentum (pf1) = 0 kg m/s (since the object comes to a stop)

Change in momentum (Δp1) = pf1 - p1 = 0 kg m/s - 10 kg m/s = -10 kg m/s

Case 2:

Initial momentum (p2) = mass (m) * velocity (v) = 1 kg * 10 m/s = 10 kg m/s

Final momentum (pf2) = 0 kg m/s (since the object comes to a stop)

Change in momentum (Δp2) = pf2 - p2 = 0 kg m/s - 10 kg m/s = -10 kg m/s

Both cases have the same change in momentum (-10 kg m/s).

Now, let's determine the impulse for each case. Impulse is defined as the product of the force applied to an object and the time interval over which it is applied.

Since the force is not explicitly given, we cannot determine the exact value of impulse for each case. However, we can still compare them based on the given information.

Case 1:

Time interval (Δt1) = 1 s

Case 2:

Time interval (Δt2) = 10 s

Since the time interval for case 2 (10 s) is greater than the time interval for case 1 (1 s), the impulse in case 2 is expected to be greater, assuming the force acting on the object is the same in both cases.

Therefore, the conclusion is:

Case 1 had the same change in momentum (-10 kg m/s) as case 2, but case 2 had the bigger impulse due to the longer time interval over which the object came to a stop.

## To determine which case had the bigger change in momentum, we can use the formula for momentum, which is given by:

Momentum (p) = mass (m) * velocity (v)

In both cases, the mass of the object is given as 1 kg. And since the object comes to a stop, the final velocity (vf) in both cases is 0 m/s.

In case 1, the object comes to a stop in 1 second. We can calculate the initial velocity (vi) using the formula for average velocity:

Average velocity (vavg) = (vf + vi)/2

Since vf = 0, we have:

vavg = (0 + vi)/2

vi = 2 * vavg = 2 * 10 m/s = 20 m/s

The change in momentum for case 1 is:

Change in momentum = p_final - p_initial = 0 - (1 kg * 20 m/s) = -20 kg*m/s

In case 2, the object comes to a stop in 10 seconds. Using the same formula for average velocity, we can calculate the initial velocity:

vavg = (0 + vi)/2

vi = 2 * vavg = 2 * 10 m/s = 20 m/s

The change in momentum for case 2 is:

Change in momentum = p_final - p_initial = 0 - (1 kg * 20 m/s) = -20 kg*m/s

Therefore, in both cases, the change in momentum is the same, resulting in -20 kg*m/s.

Now let's consider the impulse. Impulse (J) is given by the formula:

Impulse (J) = force (F) * time (t)

In both cases, we don't have the force explicitly provided. However, we can calculate the force using Newton's second law, which states:

Force (F) = mass (m) * acceleration (a)

Since the object comes to a stop, the final velocity (vf) is 0 m/s in both cases, so the acceleration (a) can be calculated as:

a = (vf - vi)/t

In case 1, the time (t) is given as 1 second. We substitute the values into the equation:

a = (0 - 10 m/s) / 1 s = -10 m/s^2

The force (F) is:

F = 1 kg * -10 m/s^2 = -10 N

In case 2, the time (t) is given as 10 seconds. Again, we substitute the values:

a = (0 - 10 m/s) / 10 s = -1 m/s^2

The force (F) is:

F = 1 kg * -1 m/s^2 = -1 N

Comparing the forces, we can see that the force in case 1 (-10 N) is greater than the force in case 2 (-1 N).

Therefore, in terms of impulse:

In case 1: Impulse = Force * Time = -10 N * 1 s = -10 N*s

In case 2: Impulse = Force * Time = -1 N * 10 s = -10 N*s

Both cases have the same impulse, resulting in -10 N*s.

In conclusion:

- Both cases have the same change in momentum (-20 kg*m/s).

- Both cases have the same impulse (-10 N*s).