please show all equations. Question: A bullet of mass 0.010 kg, moving horizonta

Question

please show all equations.

Question: A bullet of mass 0.010 kg, moving horizontally, strikes a stationary block of wood of mass 1.5 kg that is suspended as a pendulum. The bullet lodges in the wood, and together they swing upward a vertical distance of 0.40 m a) For the block/bullet combination, calculate: i. the potential energy change at maximum height; ii. the velocity and the momentum just after the collision. b) Determine the velocity and the kinetic energy of the bullet before the collision c) How much energy is dissipated as heat during the collision? 0.010kg eir -040m d) If it takes a time of 5.0 ms for the bullet to imbed in the wood, what is the average force that the bullet exerts on the block? 1.5 kg

Explanation / Answer

here,

mass of bullet , m = 0.01 kg

mass of block , M = 1.5 kg

vertical distance , h = 0.4 m

a)

the potential energy change , PE = (M + m ) * g * h

PE = ( 1.5 + 0.01 ) * 9.81 * 0.4 J

PE = 5.93 J

let the velocity be v

equating the energies

0.5 * (m + M) * v^2= (m + M) * g * h

v= sqrt(2 * 9.81 * 0.4) = 2.8 m/s

momentum , Pf = (m + M) * v = 4.23 kg.m/s

b)

let the velocity before collison be u

using conservation of momentum

m * u = Pf

0.01 * u = 4.23

u = 423 m/s

initial kinetic energy , KEi = 0.5 * m * u^2

KEi = 894.6 J

c)

the energy dissapated , E = KEi - PEf

E = 894.6 - 5.93 J = 888.7 J

d)

time taken , t = 5 ms = 0.005 s

the average force , F = m * u /t

F = 0.01 * 423 /0.005 N

F = 846 N  

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

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