<span>We know that the momentum keeps constant in a inelastic collisions, so the product of mass and speed do not change:
m1 * v1 + m2 * v2 = m * v
1 * 1 + 5 * 0 = (1 + 5) * v
1 = 6 * v
v = 1/6 m/s
So the final speed of the 6 kg chunk will travel at 0.167 m/s</span>
Since the direction of the force and the direction of the path is perpendicular, the person is not doing any physical work.
Answer:
2.667m/s to the north and 3.333 m/s to the west
Explanation:
According to law of momentum conservation, the total momentum should be conserved before and after the explosion.
Before the explosion, the momentum was
0.5*2 = 1 kg m/s to the west
Therefore the total momentum after the explosion should be the same horizontally and vertically.
Vertically speaking, it was 0 before the explosion. After the explosion:
0.2*4 + 0.3v = 0
0.3v = -0.8
v = -0.8/0.3 = -2.667 m/s
So the vertical component of the 0.3kg piece is 2.667m/s to the north
Horizontally speaking, since the 0.2kg-piece doesn't move west or east post-explosion:
0.2*0 + 0.3V = 1
0.3V = 1
V = 1/0.3 = 3.333 m/s
So the horizontal component of the 0.3kg piece is 3.333 m/s to the west
