/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 75 Measuring the speed of a bullet.... [FREE SOLUTION] | 91Ó°ÊÓ

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Measuring the speed of a bullet. A bullet of mass m is fired horizontally into a wooden block of mass M lying on a table. The bullet remains in the block after the collision. The coefficient of friction between the block and table is m, and the block slides a distance d before stopping. Find the initial speed v0 of the bullet in terms of M, m, m, g, and d.

Short Answer

Expert verified
The initial speed \( v_0 \) of the bullet can be calculated in terms of M, m, \( \mu \), g, and d using the formulas mentioned above.

Step by step solution

01

Calculation of Final Velocity after Collision

Using conservation of momentum, the final velocity \( V_f \) after the bullet impacts the block can be found by using the equation : \( V_f = \frac{m \cdot v_0}{M + m} \)
02

Calculation of Decelerations due to Friction

The deceleration due to friction \( \alpha \) can be calculated using formula: \( \alpha = \mu \cdot g \) This represents the deceleration of block under the force of friction.
03

Calculation of Initial speed of bullet

With the final velocity \( V_f \) from step 1 and deceleration \( \alpha \) from step 2, We can substitute these equations into a kinematic equation to find the initial speed \( v_0 \) of the bullet, considering the fact that the block eventually stops so \( V_f^2 = v_0^2 - 2 \alpha d \). Solving the above equation we can find \( v_0 \) in terms of M, m, \( \mu \), g and d.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Conservation of Momentum
In this exercise, the principle of conservation of momentum is crucial to solving the problem. Momentum is a measure of motion and is calculated as the product of an object's mass and velocity. For a closed system, the total momentum before an event must equal the total momentum after. This is what we mean by "conservation." Here, we have a bullet and wooden block system.
  • Before collision: Only the bullet is moving, so the initial momentum is given by the bullet's mass ( m ) multiplied by its velocity ( v_0 ).
  • After collision: Both bullet and block move together as one mass ( M + m ), which is known as a completely inelastic collision. The shared velocity is what we calculate using momentum conservation.
This results in the equation for the final velocity of the system:\[V_f = \frac{m \cdot v_0}{M + m}\]This tells us how fast the combined mass is moving just after the bullet is embedded in the block.
Kinematics
Kinematics is that branch of physics that deals with motion without regarding forces or masses. Here, it helps us understand how objects move along different paths.Once we have the block and bullet moving together, their movement is affected by friction, hence we need kinematics to describe their motion until they stop.A key equation we use is:\[V_f^2 = v_0^2 - 2 \alpha d\]Here:- \( V_f \) is the final velocity, which is zero when the block stops.- \( v_0 \) in this context is initially the combined velocity we calculated after the bullet hits the block.- \( \alpha \) is the deceleration due to friction.- \( d \) is the distance over which the block slides.This equation links the velocity, distance, and deceleration, shedding light on how quickly the block-bullet system comes to rest.
Friction and Deceleration
Friction is the force that resists the relative motion of solid surfaces. It plays a significant role in this exercise as it acts between the block and the table when the block is sliding to a stop.The coefficient of friction ( \mu ) is a dimensionless number that represents how much friction there is between the two surfaces in contact. The value of \mu we use depends on the roughness of the surfaces, in this case, the wood block and the table.Deceleration due to friction is calculated by:\[\alpha = \mu \cdot g\]This equation shows us that the deceleration is a product of the frictional force per unit mass ( \mu ) and gravity ( g ). This means that the rougher the surface or the stronger the gravitational field, the greater the deceleration.Understanding these concepts helps us determine how and why the block comes to a stop and allows us to work backwards mathematically to find the bullet's initial speed.

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Most popular questions from this chapter

A high-speed photograph of a club hitting a golf ball is shown in Figure \(6.3\). The club was in contact with a ball, initially at rest, for about \(0.0020 \mathrm{~s}\). If the ball has a mass of \(55 \mathrm{~g}\) and leaves the head of the club with a speed of \(2.0 \times 10^{2} \mathrm{ft} / \mathrm{s}\), find the average force exerted on the ball by the club.

A 0.30-kg puck, initially at rest on a frictionless horizontal surface, is struck by a 0.20-kg puck that is initially moving along the x-axis with a velocity of 2.0 m/s. After the collision, the 0.20-kg puck has a speed of 1.0 m/s at an angle of u 5 53° to the positive x-axis. (a) Determine the velocity of the 0.30-kg puck after the collision. (b) Find the fraction of kinetic energy lost in the collision.

A 2000 -kg car moving east at \(10.0 \mathrm{~m} / \mathrm{s}\) collides with a \(3000-k g\) car moving nouth. The cars stick together and move as a unit after the collision, at an angle of \(40.0^{\circ}\) north of east and a speed of \(5.22 \mathrm{~m} / \mathrm{s}\). Find the speed of the \(3000-\mathrm{kg}\) car before the collision.

Two blocks collide on a frictionless surface. After the collision, the blocks stick together. Block A has a mass M and is initially moving to the right at speed v. Block B has a mass 2M and is initially at rest. System C is composed of both blocks. (a) Draw a force diagram for each block at an instant during the collision. (b) Rank the magnitudes of the horizontal forces in your diagram. Explain your reasoning. (c) Calculate the change in momentum of block A, block B, and system C. (d) Is kinetic energy conserved in this collision? Explain your answer. (This problem is courtesy of Edward F. Redish. For more such problems, visit http://www.physics.umd.edu/perg.)

Two objects of masses m and 3m are moving toward each other along the x-axis with the same initial speed v0. The object with mass m is traveling to the left, and the object with mass 3m is traveling to the right. They undergo an elastic glancing collision such that m is moving downward after the collision at right angles from its initial direction. (a) Find the final speeds of the two objects. (b) What is the angle u at which the object with mass 3m is scattered?

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