A ball of mass 0.440 kg moving east (+x direction) with a speed of 3.80 m/s collides head-on with a 0.600 kg ball at rest. If the collision is perfectly elastic, what will be the speed and direction of each ball after the collision?
To solve this problem, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision should be equal to the total momentum after the collision.
Step 1: Calculate the initial momentum of Ball A (mass = 0.440 kg) before the collision.
Momentum (initial) = mass × velocity
Momentum (initial) = 0.440 kg × 3.80 m/s
Step 2: Calculate the initial momentum of Ball B (mass = 0.600 kg) before the collision.
Momentum (initial) = mass × velocity
Momentum (initial) = 0.600 kg × 0 m/s (since it is at rest)
Step 3: Calculate the total initial momentum of the system.
Total initial momentum = Momentum (Ball A) + Momentum (Ball B)
Step 4: Apply the conservation of momentum to determine the total momentum after the collision.
Total final momentum = Total initial momentum
Step 5: Calculate the individual final velocities of the balls.
Ball A:
Momentum (final) = mass × velocity (final)
Substitute the values and solve for velocity (final) of Ball A.
Ball B:
Momentum (final) = mass × velocity (final)
Substitute the values and solve for velocity (final) of Ball B.
Step 6: Determine the direction of motion for each ball.
Since Ball A was initially moving in the positive x direction and Ball B was initially at rest, Ball A will continue moving in the positive x direction after the collision, while Ball B will move in the negative x direction.
By following these steps, we can find the final velocities and directions of motion for each ball.
To answer this question, we can use the principles of conservation of momentum and kinetic energy.
The conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision.
Let's denote the initial velocity of the first ball (moving east) as v1, and the initial velocity of the second ball (at rest) as v2.
Before the collision:
- Ball 1: mass (m1) = 0.440 kg, velocity (v1) = 3.80 m/s
- Ball 2: mass (m2) = 0.600 kg, velocity (v2) = 0 m/s (at rest)
The total momentum before the collision is:
P_initial = m1 * v1 + m2 * v2
P_initial = (0.440 kg) * (3.80 m/s) + (0.600 kg) * (0 m/s) = 1.672 kg·m/s
After the collision, the velocity of each ball can be denoted as v1' and v2', respectively.
Using the conservation of momentum, we have:
P_initial = P_final
m1 * v1 + m2 * v2 = m1 * v1' + m2 * v2'
Applying this equation to the problem, we can solve for the final velocities of both balls.
Since the collision is perfectly elastic, the total kinetic energy before the collision should equal the total kinetic energy after the collision.
The kinetic energy (K) is given by:
K = (1/2) * m * v^2
So, the total kinetic energy before the collision is:
K_initial = (1/2) * m1 * v1^2 + (1/2) * m2 * v2^2
K_initial = (1/2) * (0.440 kg) * (3.80 m/s)^2 + (1/2) * (0.600 kg) * (0 m/s)^2
K_initial = 2.618 J
After the collision, the total kinetic energy can be denoted as K_final.
Using conservation of kinetic energy:
K_initial = K_final
(1/2) * m1 * v1^2 + (1/2) * m2 * v2^2 = (1/2) * m1 * v1'^2 + (1/2) * m2 * v2'^2
We can substitute the values of the masses and velocities to solve for the final velocities.
Since our initial velocity of the second ball is zero, the equation simplifies to:
(1/2) * m1 * v1^2 = (1/2) * m1 * v1'^2 + (1/2) * m2 * v2'^2
Substituting the values:
(1/2) * (0.44 kg) * (3.80 m/s)^2 = (1/2) * (0.44 kg) * v1'^2 + (1/2) * (0.60 kg) * v2'^2
Simplifying and re-arranging the equation, we get:
(1/2) * (0.44 kg) * (3.80 m/s)^2 - (1/2) * (0.44 kg) * v1'^2 = (1/2) * (0.60 kg) * v2'^2
Solving this equation will give us the values of v1' and v2' (final velocities of each ball after the collision).
Let
u=vel. before, and
v=vel. after collision
velocities are positive in +x direction.
Momentum is conserved for elastic and non-elastic collisions:
∑mu = ∑mv
If the collision is perfectly elastic, then kinetic energies are conserved before and after collision :
∑ (1/2)mu² = ∑(1/2)mv².
Since all velocities before collision are known, there are two equations to solve for two unknowns, v1 and v2.