A marksman at rest fires a 4.00-kg gun that expels a bullet of mass 0.0140 kg with a velocity of 181 m/s. The marksman's mass is 81.0 kg. What is the marksman's velocity after firing the gun?
mc = 85kg
md = 0.0140
-(0.0140)(-181)/85
= .0298 m/s
solve for v.
v= massbullet/massmarksman * velocty bullet.
Algebra.
my answer for this one was correct its not 0.0312
Yes, your answer was correct. You were right to add the mass of gun and marksman, since they move together. This assumes the ground he is standing on exerts no horizontal force, which is not strictly true after the shot is fired
Well, well, well, looks like we have a marksman trying to shake things up! Let's see if we can calculate their velocity after firing that gun.
If we consider the conservation of momentum, we have the formula:
m1v1 + m2v2 = (m1 + m2)v
Where m1 is the mass of the marksman (81.0 kg), v1 is the initial velocity of the marksman (which we're trying to find), m2 is the mass of the bullet (0.0140 kg), v2 is the velocity of the bullet (181 m/s), and v is the final velocity of the combined system.
Let's plug in the numbers and solve the equation:
(81.0 kg)(v1) + (0.0140 kg)(181 m/s) = (81.0 kg + 0.0140 kg)(v)
Rewriting that, we get:
81.0 kg * v1 + 0.0140 kg * 181 m/s = 81.0140 kg * v
Now, let's do some math magic and solve for v1:
v1 = (81.0140 kg * v - 0.0140 kg * 181 m/s) / 81.0 kg
After plugging in the numbers, the final answer comes out to be:
v1 ≈ 0.0298 m/s
So, the marksman's final velocity after firing the gun is approximately 0.0298 m/s. Looks like the recoil from that gun didn't knock them off their feet, but it certainly made a small difference!
To find the marksman's velocity after firing the gun, we can use the principle of conservation of momentum. This principle states that the total momentum before an event is equal to the total momentum after the event, as long as no external forces act on the system.
In this case, the system consists of the marksman and the gun. Initially, both the marksman and the gun are at rest, so their total momentum is zero. After firing the gun, the bullet moves forward with a certain velocity, and we need to find the velocity at which the marksman moves backward.
Let's denote the marksman's initial velocity as v1 and the marksman's final velocity as v2. The bullet's mass is md = 0.0140 kg, the bullet's velocity is vb = 181 m/s, and the marksman's mass is mc = 81.0 kg.
To solve the problem, we can use the equation:
(mass of the bullet * velocity of the bullet) + (mass of the marksman * initial velocity of the marksman) = (mass of the marksman * final velocity of the marksman)
Substituting the given values into the equation:
(0.0140 kg * 181 m/s) + (81.0 kg * 0 m/s) = (81.0 kg * v2)
Simplifying the equation:
2.534 kg*m/s = 81.0 kg * v2
Dividing both sides of the equation by 81.0 kg:
v2 = 2.534 kg*m/s / 81.0 kg
v2 ≈ 0.03127 m/s
Therefore, the marksman's velocity after firing the gun is approximately 0.03127 m/s in the opposite direction of the bullet's velocity.
Wouldn't it be
massbullet*velocitybullet=massmarksman*v
v then would be something different than your calcs.