Physics On a functionless horizontal surface, block A of mass 2.00 kg slides to
ID: 2278902 • Letter: P
Question
Physics
On a functionless horizontal surface, block A of mass 2.00 kg slides to the right at 15.0 m/s and block B of mass 4.00 kg slides in the same direction at 3.00 m/s. A massless spring of spring constant k = 1.120 N/m is attached on the back of block B. The masses will have an interaction via the spring, and will move apart afterwards. Find the maximum compression distance, x, of the spring between them when block A catches up with block B. Find the final velocity of block A after the collision. Suggestion: Both energy and momentum will be conseved as a result of this "collision. " Find the total momentum of the system, and the total energy. When the spring is at maximum compression the two masses move, for a brief instant, as a single mass whose velocity you can find. The kinetic energy of the two masses at this instant will be seen to be less than the total energy of the system. Where does the remaining energy go?Explanation / Answer
a) initial energy = 243 Jolue
conserving momentum the veocity whe both mass will be same be v
v*(4+2) = 15*2 + 4*3
v= 7 m*s
Energy at that time = 0.5* 6* 7^2 = 147 Joule
The remaining energy is stored in spring
0.5* 1120 * x^2 = 243-147 = 96
x = 0.414 m
b) This is elastic collision
e=1 , Veocity of aproach = velocity of sepration
v2- v1 = 15-3 = 12
Using consevation of momentum
m2v2 + m1 v1 = 4*3 +15*2 = 42
4v2+2v1 = 42
v2 = 12+v1
4(12+ v1) + 2v1 =42
6 v1 = 42-48 = -6
v1 = -1 m/s or 1 m/s to the left
v2 = 11 m/s to the right