Physics Help Forum Objects Thrown From Moving Vehicle

 Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum

 Oct 8th 2009, 09:59 AM #1 Junior Member   Join Date: Apr 2008 Posts: 6 Objects Thrown From Moving Vehicle Hi, Here's my homework problem: A car starts with constant acceleration a = 2m/s^2 at t = 0. Two coins are released from the car at t = 3 & t = 4 respectively. Each coin takes 1 second to fall on ground. Then what will the distance between the two coins be? [Assume the coins stick to the ground upon impact.] The options given are: (A) 9m (B) 7m (C) 15m (D) 2m I'm rather lost. However, I did make an attempt, and this is the lines along which I was thinking: I assumed that the coins would be carrying the same velocity as that of the car at the point at which they were thrown (Inertia). I also think that the velocities of the coins will decrease. One thing we know is that there is no acceleration acting upon them once they are separated from the car. Is that correct? Can I assume that the coins indeed do de-accelerate for that 1 second? I don't think you have to take gravity in to consideration at all, because they generally tend to specify such things. Cheers.
 Oct 8th 2009, 11:55 AM #2 Senior Member   Join Date: Sep 2009 Location: india Posts: 409 i found the ans c)15 metre. method- objects dropped from moving frames pick up velocity of the frame at the instant it was dropped but not not the acceleration of frame.so let my observation starts from the instant first stone was dropped. first stone picked velocity=0+2(3)=6m/s.it moves in a parabola and the horizontal distance it covers is v*t=6*1=6metres.before second stone was dropped, car moves (1/2*2*(4^2-3^2)=7metres.at that instant the second stone pickes vel=2*4=8m/s and thus horizontal dist traversed =8metres. so diff between two stones after they have landed=6+(7-6)+8=15metres. have i got it right?
 Oct 8th 2009, 01:29 PM #3 Junior Member   Join Date: Apr 2008 Posts: 6 I didn't know that the horizontal distance in a parabolic path is v*t. But I never thought along those lines anyway. And yeah, this looks right to me. At least it's encouraging that I got the 1st bit right (the part where the coins pick up the velocity of the frame). Thank you for the help r.samanta. Cheers.
 Oct 8th 2009, 01:38 PM #4 Senior Member   Join Date: Sep 2009 Location: india Posts: 409 actually, i used it to mean that the motion is not straight line. but you see, since coin picks up only vel but no acc, so its horizontal eq of motion will be x=ut+1/2 a t^2=ut since horizontal acc is zero.vertical direction will have acc g downward but that wont affect horizontal drift.
 Oct 8th 2009, 01:42 PM #5 Senior Member   Join Date: Sep 2009 Location: india Posts: 409 request please have a look at the thread on 'strange problem-rolling wheel' that i have recently posted in advanced mechanics section. feel free to express your views about it.

 Tags moving, objects, thrown, vehicle

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