How Fast Does the Block Move After the Bullet Embeds?

[Spraypaint]

Homework Statement

A rifle bullet of mass 7.2 grams strikes and embeds itself in a block with mass 0.819 kg that rests on a frictionless horizontal surface and is attached to a coil spring. The impact compresses the spring 17 centimeters. Calibration of the spring shows that a force of 0.68 Newtons is required to compress the spring 0.31 cm. Find the magnitude of the block’s velocity just after impact. Give your answer in m/s to the first decimal place.

Homework Equations

The Attempt at a Solution

I have looked at at least 4 other posts on this forum about this question, but I still cannot quite get it.

m1v1 = (m1+m2)v2

Using some dimensional analysis, I get that force is equal to some 37 N. Multiply this with the 17 cm, and you get the work done on the spring. This is equal to .5 * m * v1^2. Solve this for V1, plug back into the momentum equation to get V2. Is this not correct because my answer differs from the solution.

 

[Spraypaint]

Ok, so I finally found the solution, and it's something pretty stupid that I never thought of. k the spring constant is given, then everything else is simple.

 

[nlightner]

I would like to point out that it is important to include all necessary information in the problem statement, such as the units for the given values. This will help ensure that the correct equations and calculations are used.

Based on the information provided, we can use the conservation of momentum equation, m1v1 = (m1+m2)v2, to solve for the magnitude of the block's velocity just after impact.

First, we need to convert the given values to SI units. The mass of the bullet is 7.2 grams, or 0.0072 kg. The mass of the block is 0.819 kg. The displacement of the spring is 17 cm, or 0.17 m. The calibration force is 0.68 Newtons.

Using the given calibration force and displacement, we can calculate the spring constant, k, as 0.68 N / 0.31 cm = 2.19 N/cm.

Next, we can use the work-energy theorem, W = 0.5 * k * x^2, to calculate the work done on the spring by the impact of the bullet. Plugging in the values, we get W = 0.5 * 2.19 N/cm * (0.17 m)^2 = 0.0327 J.

Now, we can use the work-energy theorem again to calculate the kinetic energy of the block after the impact, as the work done by the bullet is equal to the change in kinetic energy of the block. This gives us 0.0327 J = 0.5 * 0.819 kg * v^2. Solving for v, we get v = 0.8 m/s.

Therefore, the magnitude of the block's velocity just after impact is 0.8 m/s.