A 13.5-g bullet is fired into a block of wood at 265 m/s. The block is attached

Question

A 13.5-g bullet is fired into a block of wood at 265 m/s. The block is attached to a spring that has a spring constant of 205 N/m. The block and bullet continue to move, compressing the spring by 35.0 cm before the whole system momentarily comes to a stop. Assuming that the surface on which the block is resting is frictionless, determine the mass of the wooden block. A 13.5-g bullet is fired into a block of wood at 265 m/s. The block is attached to a spring that has a spring constant of 205 N/m. The block and bullet continue to move, compressing the spring by 35.0 cm before the whole system momentarily comes to a stop. Assuming that the surface on which the block is resting is frictionless, determine the mass of the wooden block.

Explanation / Answer

Using momentum conservation:

Pi = Pf

m1v1 + m2v2 = (m1 + m2)*V

V = speed of block and bullet

V = (m1v1 + m2v2)/(m1 + m2)

V = (0.0135*265 + m2*0)/(0.0135 + m2)

V = 3.5775/(0.0135 + m2)

After that bullet + block system compresses the spring, for that part usng energy conservation

KEi + PEi = KEf + PEf

0.5*M*V^2 + 0 = 0 + 0.5*k*x^2

V = sqrt (k*x^2/M)

V = sqrt (205*0.35^2/(0.0135 + m2))

from both equation

3.5775^2/(0.0135 + m2)^2 = 205*0.35^2/(0.0135 + m2)

3.5775^2/(205*0.35^2) = 0.0135 + m2

0.5096 = 0.0135 + m2

m2 = 0.5096 - 0.0135

m2 = 0.4961 kg = 496.1 gm

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.