*Please disregard anything written on it as they are wrong. A 10 kg bomb with so

Question

*Please disregard anything written on it as they are wrong.

A 10 kg bomb with some initial velocity exploded into two fragments: nrii of mass 3.0 kg and m2 of mass 7.0 kg. Immediately after the explosion, m1 moved at 40 m/s in a direction 60 degree South of East, while m2 moved at 50 m/s in a direction 37 degree North of East. Given this situation, find the initial velocity of the bomb. Follow the steps below: Find the East-West ("horizontal") component of the bomb's velocity before the explosion. Find the North-South ("vertical") component of the bomb's velocity before the explosion. Obtain the overall velocity vector for the bomb before the explosion (find magnitude and direction). Draw a simple vector diagram to illustrate your work.

Explanation / Answer

let
M=10 kg, velocity v

m1=3kg, v1=40 m/sec, theta1=60 degrees

m2=7kg , v2=50 m/sec, theta2=37 degrees

by using conservation of momentum,

M*V=m1v1+m2v2

a)

horizontal dirction:

M*vx=m1*v1*cos(theta1)+m2*v2*cos(theta2)

10*vx=3*40*cos(60)+7*50*cos(37)

=====>

vx=33.95 m/sec

b)

vertical dirction:

M*vy=-m1*v1*sin(theta1)+m2*v2*sin(theta2)

10*vy=-3*40*sin(60)+7*50*sin(37)

=====>

vy=106.71 m/sec

c)

v=(vx)i+(vy)j

v=(33.95)i+(106.71)j

magnitude of v=sqrt(33.95^2+106.71^2)

=111.98 m/sec

and

to find the direction

use tan(theta)=vy/vx

tan(theta)=106.71/33.95

====>

theta=72.35 degrees

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