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The expanding gases that leave the muzzle of a rifle also contribute to the recoil. A 30 -caliber bullet has mass 0.00720 $\mathrm{kg}$ and a speed of 601 $\mathrm{m} / \mathrm{s}$ relative to the muzzle when fired from a rifle that has mass 2.80 kg. The loosely held rifle recoils at a speed of 1.85 $\mathrm{m} / \mathrm{s}$ relative to the earth. Find the momentum of the propellant gases in a coordinate system attached to the earth as they leave the muzzle of the rifle.

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$p_{g}=0.866 \mathrm{kg} \cdot \mathrm{m} / \mathrm{s}$ in the direction of the bullet

Physics 101 Mechanics

Chapter 8

Momentum, Impulse, and Collisions

Moment, Impulse, and Collisions

Hope College

University of Sheffield

McMaster University

Lectures

04:30

In classical mechanics, impulse is the integral of a force, F, over the time interval, t, for which it acts. In the case of a constant force, the resulting change in momentum is equal to the force itself, and the impulse is the change in momentum divided by the time during which the force acts. Impulse applied to an object produces an equivalent force to that of the object's mass multiplied by its velocity. In an inertial reference frame, an object that has no net force on it will continue at a constant velocity forever. In classical mechanics, the change in an object's motion, due to a force applied, is called its acceleration. The SI unit of measure for impulse is the newton second.

03:30

In physics, impulse is the integral of a force, F, over the time interval, t, for which it acts. Given a force, F, applied for a time, t, the resulting change in momentum, p, is equal to the impulse, I. Impulse applied to a mass, m, is also equal to the change in the object's kinetic energy, T, as a result of the force acting on it.

05:46

The expanding gases that l…

05:07

03:00

04:53

01:47

A $5.00-\mathrm{g}$ bullet…

02:54

$55 \mathrm{~m}$ When a $.…

05:29

When a .22 -caliber rifle …

01:46

A 0.11 -g bullet leaves a …

00:54

A rifle has a mass of $4.5…

01:50

Military rifles have a mec…

in discussion, we are going to find out the momentum of the propellant gases with respect to no mm possible. Let's discuss about some concept. So we know that the linear momentum of the object is equal to the product of the mass of the object and it's velocity and also that the conservation of linear momentum is stated that the initial linear momentum of the system is required to the final linear momentum of the system. So actually we discuss this because we have to find out the momentum of the propellant gases so long here, Rhonda given data, high speed of the bullet which is really active without muslim is given as V. We um equal do 601 m per second and that the coil speed of the rifle with respect to ground is given as we R G equal to -1.85 m back second. No, we consider that the direction of the bullet motion. Who are some positive direction. Hands the velocity well all it with respect uh huh. Ground, it's given us read e.g. two 601 -1.8. All right, it is equal to 599 wine 15 m part second. Now we know that expression for the momentum of the bullet with respect to ground can be written as he be equal to I am the majority of labour. We be G. Now here M V. Is the mosque of the bullet and BBg is the ferocity of correctly respect to graham. Here we substitute 0.7 to zero kg for M V And 599.15 m per second four. We busy end up about equation. Therefore we get we be equal to 0.00 7 to 0 multiply by 599.1 fight. So flammable we get BB it was too 4.3 139 kilogram meter. But second now similarly the expression for the momentum of the life will with respect to ground can be written as we are equal. Do I'm marty blob I V R G. Now here Emma is the velocity mask of the rifle and we are G. Is the required the spirit of the rifle to the respect to. Yeah. So now we here substitute 2.80 kg who hammered and minus 1.85 m per second for VR the in the above equation. So we get we are equals two little point. It's Siegel, multiply by -1.8 right? We get we are equals two -5.18 telegram meter per second. Oh we are going to calculate the momentum of the propellant gases with respect to come. So oh, conservation uh momentum. We get that the total momentum of the system is equal to zero. That means maybe yes B r. Yes E E It was 20. Now here we substitute 4.3139 kg meter for second for T v minus 5.18 kg without me for a second For pr in the above equation to solve for p p. Therefore we get 4.3139 less minus 5.18 less the peak, it was zero. So from album riga we B equals two 0.87 the low ground meter per second, and it is directly and the election uh huh, bullet and the momentum of the proponent. Yes, it with respect to long because you 0.87 kg meter per second and the direction of guesses is same as the direction of bullets.

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