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Chapter 7: Problem 50

A 2.0-kg object is at rest on a perfectly frictionless surface when it is hit by a 3.0-kg object moving at \(8.0 \mathrm{m} / \mathrm{s}\) If the two objects are stuck together after the collision, what is the speed of the combination?

Short Answer

Expert verified
Answer: The velocity of the combined objects after the collision is 4.8 m/s.

Step by step solution

01

Calculate the initial momentum of the system

To calculate the initial momentum of the system (before the collision), we need to consider the momentum of both objects. Since the stationary object is not moving, it doesn't contribute to the initial momentum. For the moving object: momentum_1 = mass_1 * velocity_1 where mass_1 = 3.0 kg (mass of the moving object) velocity_1 = 8.0 m/s (velocity of the moving object) momentum_1 = 3.0 kg * 8.0 m/s = 24 kg*m/s The initial momentum of the system is 24 kg*m/s.
02

Apply the conservation of linear momentum

According to the conservation of linear momentum, the initial momentum of the system should be equal to the final momentum of the system after the collision. Since both the objects are stuck together after the collision, we can treat them as one object with a total mass of (mass_1 + mass_2) and a common velocity (let's call it v_f): mass_2 = 2.0 kg (mass of stationary object) momentum_final = (mass_1 + mass_2) * v_f = 24 kg*m/s (from step 1)
03

Calculate the velocity of combined objects after collision

To find the velocity of the combined objects after the collision, we need to solve the equation obtained in step 2 for v_f: v_f = momentum_final / (mass_1 + mass_2) v_f = 24 kg*m/s / (3.0 kg + 2.0 kg) v_f = 24 kg*m/s / 5.0 kg v_f = 4.8 m/s The velocity of the combined objects after the collision is 4.8 m/s.

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Most popular questions from this chapter

Two objects with masses \(m_{1}\) and \(m_{2}\) approach each other with equal and opposite momenta so that the total momentum is zero. Show that, if the collision is elastic, the final speed of each object must be the same as its initial speed. (The final velocity of each object is not the same as its initial velocity, however.)
A stationary 0.1-g fly encounters the windshield of a \(1000-\mathrm{kg}\) automobile traveling at \(100 \mathrm{km} / \mathrm{h} .\) (a) What is the change in momentum of the car due to the fly? (b) What is the change of momentum of the fly due to the car? (c) Approximately how many flies does it take to reduce the car's speed by \(1 \mathrm{km} / \mathrm{h} ?\)
A 6.0 -kg object is at rest on a perfectly frictionless surface when it is struck head-on by a 2.0 -kg object moving at \(10 \mathrm{m} / \mathrm{s} .\) If the collision is perfectly elastic, what is the speed of the 6.0-kg object after the collision? [Hint: You will need two equations.]
A 0.010-kg bullet traveling horizontally at \(400.0 \mathrm{m} / \mathrm{s}\) strikes a 4.0 -kg block of wood sitting at the edge of a table. The bullet is lodged into the wood. If the table height is \(1.2 \mathrm{m},\) how far from the table does the block hit the floor?
Prove Eq. $(7-13) \Sigma \overrightarrow{\mathbf{F}}_{\mathrm{ext}}=M \overrightarrow{\mathbf{a}}_{\mathrm{CM}} .\( [Hint: Start with \)\Sigma \overrightarrow{\mathbf{F}}_{\mathrm{cxt}}=\lim _{\Delta t \rightarrow 0}(\Delta \overrightarrow{\mathbf{p}} / \Delta t),\( where \)\Sigma \overrightarrow{\mathbf{F}}_{\mathrm{ext}}$ is the net external force acting on a system and \(\overrightarrow{\mathbf{p}}\) is the total momentum of the system.]
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