Two bullets (m1=60.0g and m2=40.0g) move horizontally toward a stationary wooden

Question

Two bullets (m1=60.0g and m2=40.0g) move horizontally toward a stationary wooden block (m3=4.0kg) with initials speeds of 2v and y, respectively. Assume that the bullets strike the block at exactly the same instant and that v=400.0m/s. The block initially rests on a frictionless surface, and will climb the hill to a height h after the collision. For calculations pertaining to actual collisions, you may assume that all external forces are negligible (compared with the next collision force). a) CASE A: Find h if both bullets embed themselves simultaneously within the block. (4 pts) b) CASE B: Find h if the first bullet (mi) emerges with half of its original speed, while the second bullet (m2) embeds itself within the block. (3 pts) c) In each CASE above, what average force acts on the block if the collision lasts for 0.10ms? (3 pts)

Explanation / Answer

a)

Let Vf is the celocity of block hust after the bullet embeded into it.

Apply conservation of momentum

m1*2*v + m2*(-v) = (m1+m2+m3)*Vf

0.06*2*400 + 0.04*(-400) = (0.06 + 0.04 + 4)*Vf

==> Vf = (0.06*2*400 + 0.04*(-400))/(0.06 + 0.04 + 4)

= 7.8 m/s

now Apply conservation of energy

(m1+m2+m2)*g*h = (1/2)*(m1+m2+m2)*v^2

h = vf^2(2*g)

= 7.8^2/(2*9.8)

= 3.1 m

b)

Let Vf is the celocity of block hust after the bullet embeded into it.

Apply conservation of momentum

m1*2*v + m2*(-v) = m1*v + (m2+m3)*Vf

0.06*2*400 + 0.04*(-400) = 0.06*400 + ( 0.04 + 4)*Vf

0.06*2*400 + 0.04*(-400) - 0.06*400 = ( 0.04 + 4)*Vf

==> Vf = (0.06*2*400 + 0.04*(-400) - 0.06*400 )/(0.04 + 4)

= 1.98 m/s

now Apply conservation of energy

(m1+m2+m2)*g*h = (1/2)*(m1+m2+m2)*v^2

h = vf^2(2*g)

= 1.98^2/(2*9.8)

= 0.2 m

c)

in case a) Apply, Impulse = change in momentum

F*dt = m3*(v2 - v1)

F = m3*(v2-v1)/dt

= 4*(7.8 - 0)/(0.1*10^-3)

= 312000 N

in case b) Apply, Impulse = change in momentum

F*dt = m3*(v2 - v1)

F = m3*(v2-v1)/dt

= 4*(1.98 - 0)/(0.1*10^-3)

= 79200 N

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