A wagon is coasting at a speed, vA, along a straight and level road. When ten percent of the wagon's mass is thrown off the wagon, parallel to the ground and in the forward direction, the wagon is brought to a halt. If the direction in which this mass is thrown is exactly reversed, but the speed of this mass relative to the wagon remains the same, the wagon accelerates to a new speed, vB. Calculate the ratio of vB/vA.
Damon is absolutely right!
m and .1 m
original momentum
= m VA
momentum remains m Va throiughout
so
m Va = .1 m (Va+V)
and
m Va = .1 m (Va-V) + .9 m Vb
so
.1 m Va+.1 m V = .1 m Va-.1 m V+.9 m Vb
.2 m V = .9 m Vb
so
V = 4.5 Vb
but
m Va = .1 m Va + .1 m V
so
V = 9 Va
so
9 Va = 4.5 Vb
Vb/Va = 2
To solve this problem, we need to use the principle of conservation of momentum. According to this principle, the total momentum of a system before an event is equal to the total momentum after the event.
Let's assume the mass of the wagon is M and its initial speed is vA. The velocity of the wagon, which is represented by the product of its mass and speed, is given by the equation: M * vA.
When 10% of the wagon's mass (0.1M) is thrown off in the forward direction with the same speed (relative to the wagon), the momentum of the system remains conserved.
Initial momentum before mass is thrown off = Final momentum after mass is thrown off
M * vA = (M - 0.1M) * vA + (0.1M) * vA
Simplifying this equation, we get:
M * vA = 0.9M * vA + 0.1M * vA
Now, let's find the final speed of the wagon, vB. We know that after 0.1M is thrown off, the velocity of the wagon (M - 0.1M) * vB is equal to the mass of the thrown-off portion (0.1M) * vA.
So, (M - 0.1M) * vB = (0.1M) * vA
0.9M * vB = 0.1M * vA
Canceling out the mass terms, we get:
vB = (0.1/0.9) * vA
vB/vA = 0.1/0.9 = 1/9
Therefore, the ratio of vB to vA is 1/9.
No, youa are wrong. It is not
mVa = .1m(Va-V) + .9mVb
it is
mVa = (-).1m(Va+V) + .9mVb
it is not 2, it is 20/9