Blocks A (mass 4.00 kg ) and B (mass 15.00 kg , to the right of A) move on a fri
ID: 1453357 • Letter: B
Question
Blocks A (mass 4.00 kg ) and B (mass 15.00 kg , to the right of A) move on a frictionless, horizontal surface. Initially, block B is moving to the left at 0.500 m/s and block A is moving to the right at 2.00 m/s. The blocks are equipped with ideal spring bumpers. The collision is head on, so all motion before and after it is along a straight line. Let +x be the direction of the initial motion of A. A. Find the maximum energy stored in the spring bumpers. B. Find the velocity of block A when the energy stored in the spring bumpers is maximum. C. Find the velocity of block B when the energy stored in the spring bumpers is maximum. D. Find the velocity of block A after the blocks have moved apart. E. Find the velocity of block B after the blocks have moved apart. I've gotten 9.875J for Part A but I can't seem to figure out the rest.
Explanation / Answer
here,
Part A)
The maximum spring compression occurs when the velocities are the same (otherwise the spring is compressing or decompressing).
To find the common velocity, conserve momentum:
4 * 2 - 15 * 0.5 = 19 * v
v = 0.026 m/s
initial KE = 0.5 * 4 * 2^2 + 0.5 * 15 * 0.5^2
KEi = 9.88 J
final KE = 0.5 * 19 * (0.026)^2 = 6.422 * 10^-3 J
so max U = KEf - KEi
= 9.87 J (A)
B)
the velocity of A is 0.026 m/s to the right
C)
the velocity of B is 0.026 m/s to the right
D)
using conservation of momentum:
4 * 2 - 15 * 0.5 = 4 * v1 + 15 * v2 ...(1)
for v1 and v2 the final velocity of A and B, respectively.
For an elastic,
using conservation of kinetic energy
0.5 * m1*u1^2 + 0.5 * m2 * u2^2 = 0.5 * m1 * v1^2 + 0.5 * m2 * v2^2
4 * 2^2 + 15 * 0.5^2 = 4 * v1^2 + 15 * v2^2 ...(2)
from (1) and (2)
v1 = - 1.94 m/s
v2 = 0.55 m/s
velocity of block A is 1.94 m/s to the left