Answer:
Answer: -1.4832N. Using the impulse-momentum theorem, which states that the impulse experienced by an object is equal to the change in momentum of the object.
The impulse experienced by the bullet while traveling down the barrel is equal to the change in momentum.
The momentum of an object is given by the product of its mass and velocity:
Momentum = mass \times velocityMomentum=mass×velocity
Now, force (FF) is defined as the rate of change of momentum (\Delta pΔp) with respect to time (Delta tΔt). In this case, the time taken for the bullet to travel down the barrel is the time of firing, which is very short and not provided. However, we can simplify this by assuming the time taken for the bullet to leave the barrel is very short, almost instantaneous, making \Delta tΔt very small.
Since Delta tΔt is very small, this equation can be simplified to:
F≈−1.4832N
The negative sign indicates that the force is in the opposite direction to the motion of the bullet. So, the force exerted on the bullet while it's traveling down the barrel is approximately 1.4832- Answer: -1.4832N.
Explanation:
Answer:
310 N
Explanation:
There are two ways to solve this problem. From Newton's second law of motion, the force on the bullet is equal to its mass times its acceleration, which can be found with kinematics. Alternatively, the work-energy theorem says that the work done on the bullet is equal to its change in kinetic energy, which is half the mass times the speed.
Method 1: Newton's second law and kinematics
Given:
s = 0.77 m
u = 0 m/s
v = 322 m/s
Find: a
v² = u² + 2as
322² = 0² + 2a (0.77)
a = 67,327 m/s²
Newton's second law:
F = ma
F = (0.0046 kg) (67,327 m/s²)
F = 310 N
Method 2: Work-energy theorem
W = ΔKE
Fd = ½ mv²
F (0.77 m) = ½ (0.0046 kg) (322 m/s)²
F = 310 N