1. In the figure, block 2 (mass 1.90 kg) is at rest on a frictionless surface an

Question

1. In the figure, block 2 (mass 1.90 kg) is at rest on a frictionless surface and touching the end of an unstretched spring of spring constant 257 N/m. The other end of the spring is fixed to a wall. Block 1 (mass 1.40 kg), traveling at speed v1 = 4.30 m/s, collides with block 2, and the two blocks stick together. When the blocks momentarily stop, by what distance is the spring compressed?

2. A shell is shot with an initial velocity V0 of 19 m/s, at an angle of ?0 = 62° with the horizontal. At the top of the trajectory, the shell explodes into two fragments of equal mass (see the figure). One fragment, whose speed immediately after the explosion is zero, falls vertically. How far from the gun does the other fragment land, assuming that the terrain is level and that air drag is negligible?

Explanation / Answer

Use conservation of momentum to find the velocity immediately after impact, then use energy equations to figure the compression

momentum equations

m1u1 + m2u2 = (m1 + m2)v
v = m1u1 + m2u2 / (m1 + m2)

v = 1.40(4.30) + 1.9(0) / (1.40 + 1.90)
v = 1.82 m/s

the energy of the two masses after collision is
KE = ½mv²
which will be converted into spring potential energy
Us = ½kx²
½kx² = ½mv²
x² = mv²/k
x = (mv²/k)
x = ((1.40 + 1.90)(1.82 ²)/257)
x = 0.173 m or 17.3 cm

2) The center of mass just follows same path it was already on. The only way one fragment would fall straight down is if a huge force pushed back on it, so the other one would go farther. It would be traveling at twice the velocity that is had at the time of the explosion.

Conservation of energy applies but it was internal energy from the explosion that was converted to kinetic energy.

Use conservation of momentum and projectile motion for the rest.

vi(horizontal) = 19m/s cos(62)
vi(vertical) = 19m/s (sin62)
g = 9.8m/s^2

Time to get to top = 19m/s (sin62) / (9.8m/s^2)
1.71seconds

Max height = 1/2 gt^2 =0.5*9.8*1.71^2
14.32 meters

Horizontal displacement during that time = 19m/s(cos62) x time to get to top =19m/s(cos62) x1.71
15.25 meters



Then the horizontal velocity of the part going down will be zero and the other one will be 2(19m/s)(cos62)
17.83m/s

Time to come down = 19m/s(sin62) / 9.8m/s^2
1.71s

Horizontal displacement while coming down = 2(19m/s)(cos62) x time to come down =2(19m/s)(cos62) x1.71
30.50 meters


15.25 meters going up.
Explodes.
1/2 fragment goes straight down.
Other 1/2 fragment goes 30.50 meters on the way down.

Total distance: 45.75 meters

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