A sled of mass 5.98 kg is coasting along on a frictionless ice-covered lake at a constant speed of 1.42 m/s. A 1.98-kg book is dropped vertically onto the sled. At what speed does the sled move once the book is on it?
It moves at a speed that conserves linear momentum.
5.98*Vinitial = (5.98 +1.98)*Vfinal
Solve for Vfinal
To solve this question, we need to apply the principle of conservation of momentum. The momentum of an object is the product of its mass and velocity.
Initially, when the sled is coasting along, the momentum of the sled and the momentum of the book are both zero because the sled and book are at rest.
When the book is dropped vertically onto the sled, it exerts a downward force on the sled. According to Newton's third law of motion, there is an equal and opposite reaction, so the sled exerts an upward force on the book. This upward force changes the book's downward velocity to zero, and as a result, the book comes to rest.
However, the sled must gain some downward velocity to conserve momentum.
Let's denote the final velocity of the sled as V. According to the conservation of momentum:
(initial momentum of sled) + (initial momentum of book) = (final momentum of sled) + (final momentum of book)
The initial momentum of the sled and book is zero because they are initially at rest. The final momentum of the book is also zero because it comes to rest. Therefore, we can write the equation as:
(0 kg*m/s) + (0 kg*m/s) = (mass of sled * final velocity) + (mass of book * 0 m/s)
Simplifying the equation, we get:
0 = (mass of sled * final velocity)
Now, substitute the given values:
0 = (5.98 kg * V)
To solve for V, divide both sides of the equation by 5.98 kg:
0 / 5.98 kg = V
Therefore, the final velocity of the sled once the book is on it is 0 m/s. It comes to rest because the downward momentum of the book is equal to the upward momentum gained by the sled.