A ballistic pendulum consists of a block of wood (mass 6.00 kg) suspended by a 0

Question

A ballistic pendulum consists of a block of wood (mass 6.00 kg) suspended by a 0.70 m long wire of negligible mass. A 1.2x 10^-2 kg bullet is fired with a speed of 380 m/s toward the block. The bullet embeds itself in the block, and the block and the embedded bullet swing to a Vertical height, h, as shown below. Find the speed of the block with the bullet in it immediately after the collision. Answer: (b) (4 pts) Compute the vertical height h to which the block with embedded bullet rises. Answer: If the bullet travels into the block for a distance d = 0.2 m before getting completely stuck, what is the average force that the block exerts on the bullet?

Explanation / Answer

a) let u is the speed of bullet with the block.

Apply meomtnum conecrvation,

m1*u1 + m2*u1 = (m1+m2)*u

1.2*10^-2*380 + 0 = (1.2*10^-2+6)*u

==> u = 1.2*10^-2*380/(1.2*10^-2+6)

= 0.76 m/s

b) h = u^2/(2*g)

= 0.76^2/(2*9.8)

= 0.029 m or 2.9 cm

c) Apply, v2^2 - v1^2 = 2*a*d

0.76^2 - 380^2 = 2*a*0.2

=> a = (0.76^2 - 380^2)/(2*0.2)

= -3.6*10^5 m/s^2

F = m*|a|

= 1.2*10^-2*3.6*10^5

= 4320 N

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