A bomb of mass M at rest explodes into three pieces in the ratio 1 : 1 : 2. The
smaller ones are thrown off in perpendicular directions with velocities 3 ms−1
and 4 ms−1. What is the velocity of the third piece after the explosion?
you need v such that
3i + 4j + 2(xi+yj) = 0
3+2x = 0
4+2y = 0
v = -3/2 i -2j
now just get the magnitude and direction.
To solve this problem, we can make use of the principle of conservation of linear momentum. According to this principle, the total momentum before the explosion is equal to the total momentum after the explosion.
Let's denote the mass of the smaller pieces as m1 and m2, and the mass of the third piece as m3. Given that the ratio of their masses is 1:1:2, we can set up the following equation:
M = m1 + m2 + m3 ----(1)
Now, let's consider the momentum of the system before the explosion. Since the bomb is at rest, the initial momentum is zero.
P_initial = 0 ----(2)
After the explosion, the smaller pieces are thrown off in perpendicular directions with velocities of 3 m/s and 4 m/s. Let's denote the velocity of the third piece as v3.
The momentum of the smaller pieces after the explosion can be calculated as follows:
P_small pieces = m1 * v1 + m2 * v2 ----(3)
On the other hand, the momentum of the third piece after the explosion can be calculated as:
P_third piece = m3 * v3 ----(4)
According to the principle of conservation of linear momentum, the total momentum after the explosion should be equal to zero since the initial momentum was zero.
P_final = 0 ----(5)
Now, we can combine equations (2), (3), (4), and (5):
0 = m1 * v1 + m2 * v2 + m3 * v3 ----(6)
We know the values of v1 and v2 (3 m/s and 4 m/s), and the ratio of the masses (1:1:2). We also know the total mass of the bomb (M). Using equation (1), we can express m1 and m2 in terms of m3:
M = m1 + m2 + m3
M = (1/4)m1 + (1/4)m1 + m3
2M = m1 + m1 + 4m3
2M = 2m1 + 4m3
M = m1 + 2m3
m1 = M - 2m3 ----(7)
Substituting equation (7) into equation (6), we can solve for v3:
0 = (M - 2m3)(3) + (M - 2m3)(4) + m3 * v3
0 = 3M - 6m3 + 4M - 8m3 + m3 * v3
0 = 7M - 14m3 + m3 * v3
Rearranging the equation, we can solve for v3:
14m3 = 7M + m3 * v3
14m3 - m3 * v3 = 7M
m3(14 - v3) = 7M
m3 = (7M) / (14 - v3)
Now, since we know that m3 = 2m1, we can substitute this expression and solve for v3:
2m1 = (7M) / (14 - v3)
2(M - 2m3) = (7M) / (14 - v3)
2M - 4m3 = (7M) / (14 - v3)
2(7M - 14m3) = 7M
14M - 28m3 = 7M
14M = 35m3
M = 35/14 * m3
M = 5/2 * m3
Now, substituting the value of M in terms of m3, we can solve for v3:
M = 5/2 * m3
35/14 * m3 = 5/2 * m3
35 = 70
14 = 4v3
v3 = 14/4
v3 = 3.5 m/s
Therefore, the velocity of the third piece, after the explosion, is 3.5 m/s.