# Ballistic pendulum problem

A bullet (m = 0.0270 kg) is fired with a speed of 99.00 m/s and hits a block (M = 2.60 kg) supported by two light strings as shown, stopping quickly. Find the height to which the block rises.
I found this by conservation of energy, the answer was .0528 m
Find the angle (in degrees) through which the block rises, if the strings are 0.280 m in length.
I tried doing this using trig.
the hypotenuse= .280 m
opposite = (.280-.0528) = .2272 m
sin^-1 (.2272/.280)= 54.2 deg.

What am I doing wrong?

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if the hypotunse is .28 and the vertical side is

.28 - .0528 = .2272

so the side opp is

sqrt(.28^2-.2272^2) = .163

so sin^-1 = .163 / .28 = 35.76 degrees

Andrew Mason
Homework Helper
Punchlinegirl said:
A bullet (m = 0.0270 kg) is fired with a speed of 99.00 m/s and hits a block (M = 2.60 kg) supported by two light strings as shown, stopping quickly. Find the height to which the block rises.
I found this by conservation of energy, the answer was .0528 m
Find the angle (in degrees) through which the block rises, if the strings are 0.280 m in length.
I tried doing this using trig.
the hypotenuse= .280 m
opposite = (.280-.0528) = .2272 m
sin^-1 (.2272/.280)= 54.2 deg.
What am I doing wrong?
I am not sure what you have done because you have not shown your work explaining your answer of .0528 m for the height.

Conservation of energy does not apply to the collision. The bullet collides with the pendulum and stops inside it. This is an inelastic collision. What is conserved in the collision? Conservation of energy applies after the collision when the bullet and pendulum bob move together. (Edit: your answer .0528 m appears to be correct).

What is the height the pendulum reaches in terms of its initial velocity (after stopping the bullet)? How is the height related to angle of swing?

AM

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