My Math Forum Collision

 Physics Physics Forum

 October 2nd, 2019, 10:23 AM #1 Member   Joined: Feb 2018 From: Iran Posts: 56 Thanks: 3 Collision A body is thrown upward with a velocity of 30m/s after 5s it explodes in two pieces, one of pieces move downward with a velocity of 5m/s find the velocity of second piece in m/s ??
 October 2nd, 2019, 10:56 AM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,588 Thanks: 1432 apply conservation of linear momentum
 October 2nd, 2019, 11:36 AM #3 Senior Member   Joined: Jun 2019 From: USA Posts: 310 Thanks: 162 I don't understand the situation. If there's an "up," that means there's gravity. Should we assume standard gravity? I assume we're neglecting drag, and that 5 m/s is the instantaneous speed of the downward-moving piece immediately after the explosion, yes? Even with these questions answered, we're missing information. To apply conservation of momentum, we need to know the mass ratio of the two pieces. If 5 % of the mass is going downward, it's going to be a lot different than if it's 95 %. Thanks from topsquark
October 2nd, 2019, 12:45 PM   #4
Newbie

Joined: Oct 2019
From: SV USA

Posts: 3
Thanks: 1

Quote:
 Originally Posted by Elize A body is thrown upward with a velocity of 30m/s after 5s it explodes in two pieces, one of pieces move downward with a velocity of 5m/s find the velocity of second piece in m/s ??
Why did it explode? Do we need to factor for explosives here?

 October 2nd, 2019, 12:58 PM #5 Senior Member   Joined: Dec 2015 From: somewhere Posts: 734 Thanks: 98 Downward velocity , $\displaystyle v=v_0 +gt$. Upward velocity , $\displaystyle v=v_0 -gt$.
October 2nd, 2019, 01:09 PM   #6
Senior Member

Joined: Sep 2015
From: USA

Posts: 2,588
Thanks: 1432

Quote:
 Originally Posted by idontknow Downward velocity , $\displaystyle v=v_0 +gt$. Upward velocity , $\displaystyle v=v_0 -gt$.
suppose the explosion caused the pieces to be accelerated by non-constant transient forces!

October 2nd, 2019, 01:12 PM   #7
Senior Member

Joined: Jun 2019
From: USA

Posts: 310
Thanks: 162

Quote:
 Originally Posted by idontknow Downward velocity , $\displaystyle v=v_0 +gt$. Upward velocity , $\displaystyle v=v_0 -gt$.
One piece will not be accelerating upward.

Gravity (and probably some amount of drag) are the only forces on the object, except during the explosion. Most explosions are fast enough to be treated as instantaneous relative to a five-second time frame, so we can model it as an instantaneous impulse.

There's still not enough information to solve the problem, however.

 October 2nd, 2019, 03:42 PM #8 Senior Member     Joined: Sep 2015 From: USA Posts: 2,588 Thanks: 1432 Sometimes with incomplete problems like this you have to go ahead and make assumptions and call them out in the answer. In this problem I'd assume that both pieces are the same size and say so. Further, they want the instantaneous velocity immediately after the explosion. So gravity and any other forces don't enter into it as they don't have any time to cause an effect. so assuming the original piece had a mass of 2m and broke into 2 pieces of 1m each 30 x 2m = -5m + vm 60=-5+v v=65 m/s
 October 2nd, 2019, 05:28 PM #9 Senior Member   Joined: Jun 2019 From: USA Posts: 310 Thanks: 162 OK, I guess whether or not gravity comes into play is a matter of punctuation. 1) A body is thrown upward with a velocity of 30 m/s. After 5 s, it explodes in two pieces. 2) A body is thrown upward, with a velocity of 30 m/s after 5 s. It explodes in two pieces.
October 2nd, 2019, 06:38 PM   #10
Senior Member

Joined: Sep 2015
From: USA

Posts: 2,588
Thanks: 1432

Quote:
 Originally Posted by DarnItJimImAnEngineer OK, I guess whether or not gravity comes into play is a matter of punctuation. 1) A body is thrown upward with a velocity of 30 m/s. After 5 s, it explodes in two pieces. 2) A body is thrown upward, with a velocity of 30 m/s after 5 s. It explodes in two pieces.
Actually I'm wrong. One does have to take into account gravity for that 5s to find out what the velocity at the time of the explosion was.

30 - 9.8 x 5 = -19 m/s

-38 = -5 + v

v = -33 m/s

 Tags collision

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post markosheehan Applied Math 0 May 9th, 2017 04:12 AM markosheehan Applied Math 1 April 24th, 2017 12:34 PM markosheehan Applied Math 0 March 15th, 2017 07:51 AM VisionaryLen Physics 6 August 17th, 2016 11:42 PM Bloodnovski Algebra 2 February 12th, 2012 03:32 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}