a mass 2.9kg object moving at a speed of 5.9m\s sticks a mass 0.88kg object initially at rest. immeditaly after the collision the 2.9kg object has a velocity of 0.99m/s directed 27 degree from its initial line of motion. that is the speed of the 0.88kg mass immeditaly after collision?
To work this, consider the conservation of motion in two perpendicular directions. You will find the solution there.
To find the speed of the 0.88 kg mass immediately after the collision, we can use the principle of conservation of momentum.
The principle of conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision, provided no external forces are acting on the system.
Let's denote the mass of the 2.9 kg object as m1 and its velocity before the collision as u1. Similarly, let's denote the mass of the 0.88 kg object as m2 and its velocity before the collision as u2.
According to the conservation of momentum, we have:
(m1 * u1) + (m2 * u2) = (m1 * v1) + (m2 * v2)
Where:
m1 = mass of the 2.9 kg object
u1 = velocity of the 2.9 kg object before the collision
m2 = mass of the 0.88 kg object
u2 = velocity of the 0.88 kg object before the collision
v1 = velocity of the 2.9 kg object after the collision
v2 = velocity of the 0.88 kg object after the collision
In this case, we know the values of m1, u1, v1, and the direction of v1. We need to find v2.
Given:
m1 = 2.9 kg
u1 = 5.9 m/s
v1 = 0.99 m/s (27 degrees from the initial line of motion)
First, let's find the horizontal and vertical components of the velocity v1:
v1_horizontal = v1 * cos(27 degrees)
v1_vertical = v1 * sin(27 degrees)
Next, we can find the horizontal and vertical components of v2 by applying conservation of momentum:
(m1 * u1) = (m1 * v1_horizontal) + (m2 * v2_horizontal)
0 = (m1 * v1_vertical) + (m2 * v2_vertical)
By dividing the second equation by the first equation, we can find the ratio between the horizontal and vertical components:
(0 = (m1 * v1_vertical) + (m2 * v2_vertical)) / (m1 * u1) = (m1 * v1_horizontal) + (m2 * v2_horizontal) / (m1 * u1)
Simplifying this equation, we get:
0 = v2_vertical - (m1 * v1_vertical) / (m2 * u1) * v2_horizontal
We can solve this equation to find the value of v2_vertical in terms of v2_horizontal:
v2_vertical = (m1 * v1_vertical) / (m2 * u1) * v2_horizontal
Now, we can substitute the known values and solve for v2_horizontal:
v1_horizontal = 0.99 m/s * cos(27 degrees) = 0.906 m/s
v1_vertical = 0.99 m/s * sin(27 degrees) = 0.448 m/s
m1 = 2.9 kg
m2 = 0.88 kg
u1 = 5.9 m/s
v2_vertical = (2.9 kg * 0.448 m/s) / (0.88 kg * 5.9 m/s) * v2_horizontal
v2_vertical = 0.1506 v2_horizontal
Since the horizontal components of both velocities are equal, v2_horizontal = u1 - v1_horizontal:
v2_horizontal = 5.9 m/s - 0.906 m/s = 4.994 m/s
Finally, we can substitute the value of v2_horizontal to find v2_vertical:
v2_vertical = 0.1506 * 4.994 m/s = 0.753 m/s
Therefore, the speed of the 0.88 kg mass immediately after the collision is approximately 0.753 m/s.