Three particles of masses 20g, 30g and 4...

Three particles of masses `20g`, `30g` and `40g` are initially moving along the positive direction of the three coordinate axes respectively with the same velocity of `20cm//s`. When due to their mutual interaction, the first particle comes to rest, the second acquires a velocity `(10hati+20hatk)cm//s`. What is then the velocity of the third particle?

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The correct Answer is:

A, B, C

`p_i=p_f`
`:. (20)(20hati)+30(20hatj)+40(20hatk)`
`=(20)(0)+30(10hati+20hatk)+40v`
Solving this equation we get,
`v=(2.5hati+15hatj+5hatk)cm//s`

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Three particles with masses 10, 20 and 40gm are moving with velocities 10hati, 10hatj and 10hatk m/sec respectively. If due to some interaction the first particle comes to rest and the velocity of second becomes ) (3hati+4hatj) m/sec. then the velocity of third particle after the interaction is-

`hati+hatj+5hatk`

`hatj+10hatk`

`hati+hatj+10hatk`

`hati+3hatj+10hatk`

`20hat(i) + 20hat(j) + 70hat(k)`

`10hat(i) + 20hat(j) + 8hat(k)`

`30hat(i) + 10hat(j) + 7hat(k)`

`15hat(i) + 5hat(j) + 60hat(k)`

3 m/s

4 m/s

5 m/s

zero