Question

A 0.50-kg bomb is sliding along an icy pond (frictionless surface) with a velocity of 2.0 m/s to the west. The bomb explodes into two pieces. After the explosion, a 0.20-kg piece moves south at 4.0 m/s. What are the components of the velocity of the 0.30-kg piece?

Answer 1

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

Answer 2

Answer:

x = - 3.33 i

y = 2.667 j

Explanation:

Given that;

mass of bomb $m1$ = 0.50 kg

After explosion , 0.20 kg piece moves south $m2 =$ 0.20 kg

At $v2 =$ 4.0 m/s

By using conservation of momentum

$m1 v1 + m2 v2 (-j) = (m1 + m2) v (-i)$

$0.3 v1 - 0.2* 4 j = - 0.5* 2 i$

$v1 = - 3.33 i + 2.667 j$

so  for each components of velocity $v1$ on the x and y axis are shown below in a respective order

x = - 3.33 i

y = 2.667 j