a mass m1 moves along the x-axis with velocity of magnitude v0 on a frictionless table. it strikes another mass m2 which is initially at rest. the mass m1 goes off along the the y-axis. if half the original kinetic energy is lost in the collision, with what speed and at what angle does m2 leave the point of collision?

To solve this problem, we can use the principles of conservation of momentum and kinetic energy.

Step 1: Conservation of momentum
When two objects collide, the total momentum of the system before the collision is equal to the total momentum after the collision. Therefore, we can write the equation for momentum conservation in the x-direction:

m1 * v0 = m1 * vx1 + m2 * vx2 (Equation 1)

where
m1 = mass of object 1
v0 = initial velocity of object 1 along the x-axis
vx1 = velocity of object 1 along the x-axis after collision
m2 = mass of object 2
vx2 = velocity of object 2 along the x-axis after collision

Since we're given that mass m1 goes off along the y-axis, we can assume its velocity along the x-axis after the collision (vx1) is zero. Therefore, Equation 1 becomes:

m1 * v0 = m2 * vx2 (Equation 2)

Step 2: Conservation of kinetic energy
We're also given that half of the original kinetic energy is lost in the collision. The initial kinetic energy of the system is:

KE_initial = (1/2) * m1 * v0^2 (Equation 3)

The final kinetic energy after the collision is:

KE_final = (1/2) * m1 * vy1^2 + (1/2) * m2 * vy2^2 (Equation 4)

where
vy1 = velocity of object 1 along the y-axis after collision
vy2 = velocity of object 2 along the y-axis after collision

Since half of the initial kinetic energy is lost, we have:

KE_final = (1/2) * KE_initial (Equation 5)

Substituting Equations 3 and 4 into Equation 5:

(1/2) * m1 * vy1^2 + (1/2) * m2 * vy2^2 = (1/2) * (1/2) * m1 * v0^2

Simplifying:

m1 * vy1^2 + m2 * vy2^2 = (1/4) * m1 * v0^2 (Equation 6)

Step 3: Solving the equations
We now have two equations (Equations 2 and 6) with two unknowns (vx2 and vy2). Solve these equations simultaneously to find the values of vx2 and vy2.

Equation 2: m1 * v0 = m2 * vx2
Solving for vx2:
vx2 = (m1 * v0) / m2

Substitute this expression for vx2 into Equation 6:

m1 * vy1^2 + m2 * ((m1 * v0) / m2)^2 = (1/4) * m1 * v0^2

Simplify:

m1 * vy1^2 + (m1^2 * v0^2) / m2 = (1/4) * m1 * v0^2

Multiply through by m2:

m1 * vy1^2 * m2 + m1^2 * v0^2 = (1/4) * m1 * v0^2 * m2

Expand:

m1 * vy1^2 * m2 + m1^2 * v0^2 = (1/4) * m1 * v0^2 * m2

Rearrange to solve for vy1^2:

vy1^2 = (1/4) * v0^2 - (m1 * v0^2) / (m1 + m2)

Now we have the value of vy1^2. Take the square root to find the magnitude of vy1.

Finally, to find the speed and angle at which m2 leaves the point of collision, use the magnitude of vy1 to calculate the resultant velocity (v2) of m2 using the Pythagorean theorem:

v2 = sqrt(vy1^2 + vx2^2)

The speed of m2 is |v2|, and the angle at which it leaves can be found using the inverse tangent function:

angle = arctan(vy1 / vx2)

Plug in the values to get the speed and angle at which m2 leaves the point of collision.

To solve this problem, we can use the principle of conservation of linear momentum and conservation of kinetic energy.

Let's denote the velocity of m1 after the collision as v1 and the velocity of m2 after the collision as v2.

According to the principle of conservation of linear momentum, the total momentum before the collision is equal to the total momentum after the collision:

m1 * v0 = m1 * v1 + m2 * v2 ---(1)

Now, let's consider the conservation of kinetic energy. We are given that half of the original kinetic energy is lost in the collision. Kinetic energy is given by (1/2) * m * v^2, where m is the mass and v is the velocity.

The initial kinetic energy before the collision is given by:
(1/2) * m1 * v0^2

The final kinetic energy after the collision is given by:
(1/2) * m1 * v1^2 + (1/2) * m2 * v2^2

Since half of the original kinetic energy is lost, we can set up the following equation:

(1/2) * m1 * v0^2 = (1/2) * m1 * v1^2 + (1/2) * m2 * v2^2 ---(2)

Now, from equation (1), we can express v1 in terms of v2:

v1 = (m1 * v0 - m2 * v2) / m1

Substitute this expression for v1 into equation (2):

(1/2) * m1 * v0^2 = (1/2) * m1 * [(m1 * v0 - m2 * v2) / m1]^2 + (1/2) * m2 * v2^2

Simplifying the equation:

m1 * v0^2 = (m1^2 * (v0^2 - 2 * v0 * v2 + v2^2)) / (2 * m1) + (m2 * v2^2) / 2

Multiply through by 2m1 to eliminate the denominators:

2 * m1^2 * v0^2 = m1^2 * (v0^2 - 2 * v0 * v2 + v2^2) + m2 * m1 * v2^2

Simplifying further:

2 * m1^2 * v0^2 = m1^2 * v0^2 - 2 * m1^2 * v0 * v2 + m1^2 * v2^2 + m2 * m1 * v2^2

Rearranging the terms:

m1^2 * v0^2 - m1^2 * v0^2 = (m1^2 + m2 * m1) * v2^2 - 2 * m1^2 * v0 * v2

Simplifying:

0 = (m1^2 + m2 * m1) * v2^2 - 2 * m1^2 * v0 * v2

Now, we can solve for v2. Dividing through by (m1^2 + m2 * m1):

0 = v2^2 - 2 * m1^2 * v0 * v2 / (m1^2 + m2 * m1)

Rearranging:

v2^2 = 2 * m1^2 * v0 * v2 / (m1^2 + m2 * m1)

Simplifying further:

v2^2 = 2 * m1 * v0 * v2 / (m1 + m2)

Finally, solving for v2:

v2 = (2 * m1 * v0) / (m1 + m2)

Now that we have found the speed of m2 after the collision, we need to find the angle at which m2 leaves the point of collision. Since m1 goes off along the y-axis, it means that m2 must move along the x-axis.

Therefore, we can conclude that the angle at which m2 leaves the point of collision is 0 degrees.

To summarize:
- The speed of m2 after the collision is given by v2 = (2 * m1 * v0) / (m1 + m2).
- The angle at which m2 leaves the point of collision is 0 degrees.