The glider collides elastically an another glider at rest, in this case we have the coservation of quantity of movement
m1 v1 + m2 v2 = m1V1 + m2 V2
v is the velocity before the shock, V the velocity after the shock
m1=0.22
v1=0.60m/s
m2=0.44
V1=-0.20m/s
remark that <span>0.44 kg glider is initially at rest, so v2= 0
the equation above can be written </span><span>m1 v1 = m1V1 + m2 V2 ( because v2=0 at rest)
</span><span>m1 v1 = m1V1 + m2 V2 implies V2= (m1v1 -m1V1) / m2
</span>
V2=0.22x0.60 -0.22x (-0.20) ] / 0.44= 0.4m/s
Scientists now gathered information and facts before creating a theory.
Answer:
X(t) = 9.8 *t - 4.9 * t^2
Explanation:
We set a frame of reference with origin at the hand of the girl the moment she releases the ball. We assume her hand will be in the same position when she catches it again. The positive X axis point upwards.The ball will be subject to a constant gravitational acceleration of -9.81 m/s^2.
We use the equation for position under constant acceleration:
X(t) = X0 + V0 * t + 1/2 * a *t^2
X0 = 0 because it is at the origin of the coordinate system.
We know that at t = 2, the position will be zero.
X(2) = 0 = V0 * 2 + 1/2 * -9.81 * 2^2
0 = 2 * V0 - 4.9 * 4
2 * V0 = 19.6
V0 = 9.8 m/s
Then the position of the ball as a function of time is:
X(t) = 9.8 *t - 4.9 * t^2
Answer:
It can go back to it's original shape
Explanation:
