A 83-kg fisherman in a 122-kg boat throws a package of mass m = 15 kg horizontally toward the right with a speed of vi = 4.7 m/s as in the figure below. Neglecting water resistance, and assuming the boat is at rest before the package is thrown, find the velocity of the boat after the package is thrown.
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Try using conservation of momentum
To find the velocity of the boat after the package is thrown, we can use the principle of conservation of momentum. According to this principle, the total momentum before an event is equal to the total momentum after the event, assuming no external forces act on the system.
In this case, the total momentum before the package is thrown is zero since the boat is at rest. After the package is thrown, the fisherman and the boat will start moving in the opposite direction to conserve momentum.
To calculate the velocity of the boat, we need to consider the masses and velocities of the fisherman, boat, and package.
Given:
Mass of the fisherman (m1) = 83 kg
Mass of the boat (m2) = 122 kg
Mass of the package (m3) = 15 kg
Initial velocity of the package (v3i) = 4.7 m/s
Let's denote the initial velocity of the fisherman and the boat as v1i and v2i, respectively. Since the boat is initially at rest, v2i = 0.
Applying momentum conservation, the total momentum before the package is thrown is given by:
m1*v1i + m2*v2i + m3*v3i
Since the total momentum before is zero (since the boat is at rest), we have:
m1*v1i + m3*v3i = 0
Solving for v1i, we get:
v1i = - (m3*v3i) / m1
Now that we have the initial velocity of the fisherman, we can find the final velocity of the boat using the conservation of momentum again.
The total momentum after the package is thrown is given by:
m1*v1f + m2*v2f + m3*v3f
Since the fisherman and boat have opposite velocities, we can denote the final velocity of the fisherman as -v1f.
So, the equation for the total momentum after the package is thrown becomes:
-m1*v1f + m2*v2f + m3*v3f
Since no external forces are acting, the total momentum before and after the package is thrown should be the same. Therefore:
m1*v1i + m3*v3i = -m1*v1f + m2*v2f + m3*v3f
Plugging in the values we know, we get:
83 * v1i + 15 * 4.7 = -83 * v1f + 122 * v2f + 15 * 4.7
Substituting the value of v1i from earlier, we have:
83 * [-(15 * 4.7) / 83] + 15 * 4.7 = -83 * v1f + 122 * v2f + 15 * 4.7
Simplifying the equation gives:
-v1f + 15 = -83 * v1f + 122 * v2f + 70.5
Rearranging the terms, we get:
82 * v1f + 122 * v2f = -84.5
Since v1f = -v2f (opposite direction), we can substitute v1f with -v2f:
82 * (-v2f) + 122 * v2f = -84.5
Simplifying, we find:
40 * v2f = -84.5
Dividing by 40:
v2f = -2.1125 m/s
The magnitude of the velocity of the boat after the package is thrown is approximately 2.1125 m/s, and the direction is to the left.