An object of mass 9.0 kg is moving due west with a uniform speed of 4.0 m/s. It suddenly explodes into two parts having masses of 6.0 kg and 3.0 kg. Just after the explosion, the 3.0 kg mass has a velocity of 15 m/s due west relative to the Earth's surface. What is the speed of the 6.0 kg mass relative to the Earth's surface?
*If possible, please show the process in how you got the answer!*
the momentum after = the momentum before
momentum before = 9 * 4 West + 0 North
momentum after = 3 *15 West + 6 * v West + 0 North
so
36 = 45 + 6 v West
6 v West = - 9
v West = -9/6 = -1.5 West or in other words 1.5 East
Thank you so much for your help!
You are welcome.
Given: M1 = 9kg, V1 = -4m/s, M2 = 6kg, V2 = ?, M3 = 3kg, V3 = -15m/s.
Momentum before = Momentum after
M1*V1 = M2*V2 + M3*V3
9*(-4) = 6*V2 + 3*(-15)
-36 = 6V2-45
V2 = 1.5 m/s, East.
To find the speed of the 6.0 kg mass relative to the Earth's surface, we can use the principle of conservation of momentum. According to this principle, the total momentum before the explosion is equal to the total momentum after the explosion.
First, let's find the momentum before the explosion. We know that the initial mass (m1) is 9.0 kg, and the initial velocity (v1) is 4.0 m/s. So, the initial momentum is given by:
Initial momentum = m1 * v1 = 9.0 kg * 4.0 m/s = 36 kg·m/s (due west)
Next, let's find the momentum after the explosion. We have two masses after the explosion (m2 = 6.0 kg and m3 = 3.0 kg) and their respective velocities (v2 and v3). The momentum of the 6.0 kg mass (m2) is given by:
Momentum of m2 = m2 * v2
And the momentum of the 3.0 kg mass (m3) is given by:
Momentum of m3 = m3 * v3
Since the two masses move in opposite directions, their momenta are of opposite signs.
According to the conservation of momentum principle, the total momentum before the explosion is equal to the total momentum after the explosion. Mathematically, this can be expressed as:
Initial momentum = Momentum of m2 + Momentum of m3
36 kg·m/s = (6.0 kg * v2) + (-3.0 kg * 15 m/s) [Substituting m2 with 6.0 kg and m3 with 3.0 kg, and v3 with -15 m/s (due west)]
Solving this equation will give us the value of v2, which is the velocity of the 6.0 kg mass relative to the Earth's surface.
36 kg·m/s = 6.0 kg * v2 - 45 kg·m/s
81 kg·m/s = 6.0 kg * v2
v2 = 81 kg·m/s / 6.0 kg
v2 ≈ 13.5 m/s
Therefore, the speed of the 6.0 kg mass relative to the Earth's surface is approximately 13.5 m/s.