A 2.5 kg stone is released from rest and falls toward earth. After 4s, the magnitude of its momentum is:
My work:
J=F*t
=(F *t)= 24.5N * 4s
F= 2.5 *9.81= 24.5
J=P
P= 98kg m/s
A sphere X, of mass 2kg, is moving to the right at 10m/s. Sphere Y, of mass 4 kg is moving to the left at 10m/s. The two sphere collide head on. The magnitude of the impulse of X and Y is:
Would the answer be half the magnitude of the impulse of Y on X because Y is two times greater in mass then X.
Yes on the first one.
One the second, remember that impulse is a vector, these vectors are in the opposite direction. You are correct, 20 kg*m/s to the left.
To find the magnitude of the momentum of the 2.5 kg stone falling towards Earth after 4 seconds, you can use the formula for momentum:
Momentum (P) = mass (m) * velocity (v)
However, in this case, the stone is released from rest, which means its initial velocity is zero. As the stone falls towards Earth, it gains velocity due to gravity. The acceleration due to gravity (g) is approximately 9.81 m/s^2.
To find the final velocity of the stone after 4 seconds, you can use the formula of motion:
Final velocity (v) = Initial velocity (u) + acceleration (a) * time (t)
Since the stone is released from rest, its initial velocity (u) will be zero. Plugging in the values, you get:
v = 0 + 9.81 * 4 = 39.24 m/s
Now that you have the velocity, you can find the momentum:
P = m * v = 2.5 kg * 39.24 m/s = 98.1 kg·m/s
Therefore, the magnitude of the stone's momentum after 4 seconds is 98.1 kg·m/s.
For the second question, you are correct that the magnitude of the impulse of sphere X on sphere Y would be half of the magnitude of the impulse of sphere Y on sphere X. This is because impulse is defined as the change in momentum, and since momentum is a vector, the magnitude of the impulse is the same regardless of the mass. So, if sphere Y has twice the mass of sphere X, the magnitude of the impulse of Y on X would be twice as much as the magnitude of the impulse of X on Y.