ball of mass 0.584 kg moving east (+ direction) with a speed of 3.78 m/s collides head-on with a 0.292 kg ball at rest. Assume that the collision is perfectly elastic.
To solve this problem, we will use the conservation of momentum and the conservation of kinetic energy.
A) Conservation of momentum:
The total momentum before the collision is:
p_before = m1v1 + m2v2 where m1 = 0.584 kg is the mass of the first ball (moving), v1 = 3.78 m/s is its velocity, m2 = 0.292 kg is the mass of the second ball (at rest), and v2 = 0 m/s is its velocity.
p_before = (0.584 kg)(3.78 m/s) + (0.292 kg)(0 m/s) = 2.196 kg*m/s
After the collision, the total momentum is conserved, so:
p_after = m1v1' + m2v2' where v1' and v2' are the velocities of the two balls after the collision.
We can solve for v1' by noting that the collision is head-on, so the direction of motion of the first ball will not change. Therefore, we have:
p_after = (0.584 kg)v1' + (0.292 kg)v2'
Since the collision is perfectly elastic, the kinetic energy is also conserved:
1/2m1v1^2 + 1/2m2v2^2 = 1/2m1v1'^2 + 1/2m2v2'^2
Substituting v2' = -v1' (due to conservation of momentum), we get:
1/2m1v1^2 = 1/2m1v1'^2 + 1/2m2v2'^2
Substituting the values given in the problem, we can solve for v1':
2.196 kg*m/s = (0.584 kg)v1' + (0.292 kg)(-v1') v1' = 5.89 m/s
Therefore, the speed of the 0.584-kg ball after the collision is 5.89 m/s, rounded to three significant figures.
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