A bomb of mass 6 kg initially at rest explodes in to three identical fragments. One
of the fragments moves with a velocity of10 3 / im s � , another fragment moves with
a velocity if � 10 j m/s, then the third fragment moves with a velocity of magnitude.
To find the velocity of the third fragment, we can apply the law of conservation of momentum. According to this principle, the total momentum before the explosion should be equal to the total momentum after the explosion.
Given:
Mass of the bomb (m) = 6 kg
Velocity of first fragment (v1) = 10i m/s
Velocity of second fragment (v2) = 10j m/s
Since the fragments are identical, let's assume the velocity of the third fragment be v3.
The momentum before the explosion is the sum of the momentum of the bomb:
Momentum before = Momentum of bomb = mass x velocity
P_initial = m x v_initial
The total initial momentum is:
P_initial = 6 kg x 0 m/s (as the bomb is initially at rest)
Now, let's calculate the momentum after the explosion:
Momentum after = Momentum of fragment 1 + Momentum of fragment 2 + Momentum of fragment 3
P_final = m1 x v1 + m2 x v2 + m3 x v3
Where m1, m2, and m3 are the masses of the three fragments.
As the fragments are identical and the mass of the bomb is distributed equally among them, we can assume m1 = m2 = m3 = 6 kg / 3 = 2 kg.
Therefore, the equation becomes:
P_final = 2 kg x 10i m/s + 2 kg x 10j m/s + 2 kg x v3
Since we only need to find the magnitude of the velocity of the third fragment, we can disregard the direction and only consider the magnitudes:
P_final = 20 kg·m/s + 20 kg·m/s + 2 kg·v3
P_final = 40 kg·m/s + 2 kg·v3
According to the law of conservation of momentum, the initial momentum is equal to the final momentum:
P_initial = P_final
However, the initial momentum P_initial is zero since the bomb is initially at rest, so:
0 = 40 kg·m/s + 2 kg·v3
Simplifying the equation:
2 kg·v3 = -40 kg·m/s (multiplying both sides by -1)
v3 = -40 kg·m/s / 2 kg
v3 = -20 m/s
Therefore, the magnitude of the velocity of the third fragment is 20 m/s.
To solve this problem, 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.
Before the explosion, the bomb is at rest, so the total momentum is zero. After the explosion, the three fragments will move with different velocities, but their total momentum should still be zero.
Let's denote the velocity of the third fragment as V.
The momentum of the first fragment (A) can be calculated using the formula:
pA = mAvA
where m is the mass of the fragment (which is one-third of the total mass, so m = 6 kg / 3 = 2 kg) and vA is the velocity of fragment A (which is given as 10√3 i m/s).
So, pA = (2 kg)(10√3 i m/s) = 20√3 i kg∙m/s
Similarly, the momentum of the second fragment (B) can be calculated using the formula:
pB = mBvB
where m is the mass of the fragment (which is one-third of the total mass, so m = 6 kg / 3 = 2 kg) and vB is the velocity of fragment B (which is given as 10 j m/s).
So, pB = (2 kg)(10 j m/s) = 20 j kg∙m/s
Now, the total momentum after the explosion should be zero, so we can use vector addition to find the velocity of the third fragment (V).
The total momentum after the explosion is given by:
pA + pB + pV = 0
Substituting the known values:
20√3 i kg∙m/s + 20 j kg∙m/s + pV = 0
To make the total momentum zero, the momentum of the third fragment (pV) should have the opposite direction and magnitude of the sum of the first two momenta. Therefore, we can write:
pV = -20√3 i kg∙m/s - 20 j kg∙m/s
The magnitude of the velocity of the third fragment can be found using the formula:
|V| = |pV| / mV
where mV is the mass of the fragment, which is one-third of the total mass (2 kg).
Substituting the known values:
|V| = |-20√3 i kg∙m/s - 20 j kg∙m/s| / 2 kg
To find the magnitude of the vector, we can use the Pythagorean theorem:
|V| = √((-20√3)^2 + (-20)^2) / 2 kg
Calculating the expression inside the square root:
|V| = √(720 + 400) / 2 kg
Simplifying:
|V| = √1120 / 2 kg
Finally:
|V| = √560 kg∙m/s
Therefore, the magnitude of the velocity of the third fragment is √560 kg∙m/s.
momentum is conserved. so
initial momentum is zero. So
2*V1+2*V2+2*V3=0
you are given v1, v2. so add them. Then if their sum is L i+Mj, then v3 must be -Li-Mj m/s
and magnitude is sqrt(L^2+M^2)