An explosive is fired from the ground at an angle of 53.1 degrees and a speed of

Question

An explosive is fired from the ground at an angle of 53.1 degrees and a speed of 30 m/s. When the explosive reaches its maximum height it breaks into two peices of equal mass. One of the pieces is hurled straight backwards and returns on the same path it previously took.

a. what is the speed of the explosive at the max height?

b. what are the soeeds of each piece after it breaks apart?

c. how far from the launch point does the far piece land?

d. at the time both pieces land what is the position of the center of mass of the system from the launch point?

Explanation / Answer

a)

v = 30 cos(53.1) = 18.0126 = 18.0 m/s

b)

one of the pieces is hurled straight backwards:

v1 = -18.0 m/s

conservation of momentum:

M v = (M/2) v1 + (M/2) v2

==> v = (1/2) v1 + (1/2) v2

==> 18 = (1/2) (-18) + (1/2) v2

==> v2 = 54 m/s

c)

t = 30*sin53.1/9.8 = 2.448 s

x2 = v2 t = 54 * 2.448 = 132 m

d)

x1 = v1 t = 18 * 2.448 = 44.06 m

xCM = (132+44.06)/2 = 88.1 m

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