A bullet (m = 0.0290 kg) is fired with a speed of 99.00 m/s and hits a block (M

Question

A bullet (m = 0.0290 kg) is fired with a speed of 99.00 m/s and hits a block (M = 2.40 kg) supported by two light strings as shown, stopping quickly.

(a) Find the height to which the block rises.

(b) Find the angle (in degrees) through which the block rises, if the strings are 0.330 m in length.

A bullet (m = 0.0290 kg) is fired with a speed of 99.00 m/s and hits a block (M = 2.40 kg) supported by two light strings as shown, stopping quickly. Find the height to which the block rises. Find the angle (in degrees) through which the block rises, if the strings are 0.330 m in length.

Explanation / Answer

1)use conservation of momentum to get velocity after the collision

(.029)(99)=(2.429)*v
v=1.182 m/s

Now we use conservation of energy to get the height

KE=PE
(1/2)mv2=mgh

h=v2/(2g)

h=1.1822/(2*9.8)

h=.07128 m

2)

h=L(1-cos?)

cos?=(L-h)/(L)

cos?=(.330-.07128)/.330

?=38.37 degrees

Hope that helps

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