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Chapter 7: Problem 43

A block of wood of mass \(0.95 \mathrm{kg}\) is initially at rest. A bullet of mass \(0.050 \mathrm{kg}\) traveling at \(100.0 \mathrm{m} / \mathrm{s}\) strikes the block and becomes embedded in it. With what speed do the block of wood and the bullet move just after the collision?

Short Answer

Expert verified
Answer: The final speed of the block of wood and bullet just after the collision is 5.0 m/s.

Step by step solution

01

Define initial and final momenta

Before the collision, the block of wood is at rest and the bullet is moving at a given speed. The initial momentum of the system is given by the momentum of the bullet since the wood block's momentum is zero (rest). After the collision, the block of wood and the bullet are moving together at an unknown speed, which we must find. Initial momentum: \(momentum_{initial} = momentum_{bullet}\) Final momentum: \(momentum_{final} = momentum_{wood + bullet}\)
02

Apply conservation of linear momentum

According to the conservation of linear momentum, the total momentum of the system before the collision is equal to the total momentum of the system after the collision. Therefore, we can write the equation: \(momentum_{initial} = momentum_{final}\)
03

Calculate initial momentum

First, we calculate the initial momentum of the bullet: \(momentum_{bullet} = mass_{bullet} \times velocity_{bullet}\) \(momentum_{bullet} = (0.050 \mathrm{kg}) \times (100.0 \mathrm{m/s})\) \(momentum_{bullet} = 5.0 \mathrm{kg \cdot m/s}\)
04

Calculate final momentum

Now we write the expression for the final momentum of the system, which is equal to the total mass (mass of block plus mass of bullet) multiplied by their unknown final velocity (v): \(momentum_{final} = (mass_{wood} + mass_{bullet}) \times v\)
05

Set up an equation and solve for final velocity

Now we can set up an equation for the conservation of linear momentum and solve for the final velocity: \(momentum_{initial} = momentum_{final}\) \(mass_{bullet} \times velocity_{bullet} = (mass_{wood} + mass_{bullet}) \times v\) Plugging in the values for masses and velocity: \(5.0 \mathrm{kg \cdot m/s} = (0.95 \mathrm{kg} + 0.050 \mathrm{kg}) \times v\) \(v = \frac{5.0 \mathrm{kg \cdot m/s}}{1.0\mathrm{kg}}\) \(v = 5.0 \mathrm{m/s}\)
06

State final answer

Therefore, the final speed of the block of wood and bullet together just after the collision is 5.0 m/s.

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Most popular questions from this chapter

A sports car traveling along a straight line increases its speed from $20.0 \mathrm{mi} / \mathrm{h}\( to \)60.0 \mathrm{mi} / \mathrm{h} .$ (a) What is the ratio of the final to the initial magnitude of its momentum? (b) What is the ratio of the final to the initial kinetic energy?
Two identical pucks are on an air table. Puck A has an initial velocity of \(2.0 \mathrm{m} / \mathrm{s}\) in the \(+x\) -direction. Puck \(\mathrm{B}\) is at rest. Puck A collides elastically with puck B and A moves off at $1.0 \mathrm{m} / \mathrm{s}\( at an angle of \)60^{\circ}\( above the \)x$ -axis. What is the speed and direction of puck \(\mathrm{B}\) after the collision?
A 3.0-kg body is initially moving northward at \(15 \mathrm{m} / \mathrm{s}\) Then a force of \(15 \mathrm{N},\) toward the east, acts on it for 4.0 s. (a) At the end of the 4.0 s, what is the body's final velocity? (b) What is the change in momentum during the \(4.0 \mathrm{s} ?\)
Consider two falling bodies. Their masses are \(3.0 \mathrm{kg}\) and $4.0 \mathrm{kg} .\( At time \)t=0,$ the two are released from rest. What is the velocity of their \(\mathrm{CM}\) at \(t=10.0 \mathrm{s} ?\) Ignore air resistance.
An object of mass \(3.0 \mathrm{kg}\) is projected into the air at a \(55^{\circ}\) angle. It hits the ground 3.4 s later. What is its change in momentum while it is in the air? Ignore air resistance.
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