64.8k views
3 votes
What is the speed v of the block and magnet immediately after the collision? View available hint(s) for part a v.

(A question about specific values and the collision scenario would need to be provided for options)
A) The correct calculated speed
B) An incorrect calculated speed
C) Not enough information to determine
D) None of the above

User Solidus
by
8.2k points

1 Answer

3 votes

Final answer:

The final speed of the two objects after a totally inelastic collision can be found using the conservation of momentum. Mass A, moving at 24 m/s, collides with mass B at rest, which is twice as massive. Post-collision, both objects move together at 8 m/s in the +x-direction.

Step-by-step explanation:

The question pertains to the final speed of two objects after a totally inelastic collision. The scenario given is that mass A is moving at 24 m/s in the +x-direction and mass B, which is twice as massive as mass A, is initially at rest. To find the final speed of both objects after the collision, one must apply the principle of conservation of momentum.

The key formula to use here is the initial momentum equals the final momentum, so for totally inelastic collisions, (m1 * v1) + (m2 * v2) = (m1 + m2) * v'. Given mass A (m1) is moving at 24 m/s and mass B (m2) is at rest, the final velocity (v') can be calculated as follows: (m1 * v1) / (m1 + m2) = (m1 * 24 m/s) / (m1 + 2 * m1) = 24 m/s / 3 = 8 m/s. Therefore, after the collision, both masses A and B move together with a final speed of 8 m/s in the +x-direction.

User Vrutberg
by
8.4k points