Forest UCM MnAM InElasticCol - Revision history

Foretony: /* problem 3.4 two hobos */

2014-09-15T15:40:18Z

problem 3.4 two hobos

2014-09-12T12:06:08Z

2014-09-12T12:01:24Z

2014-09-12T12:00:57Z

2014-09-12T12:00:33Z

2014-09-12T11:59:39Z

problem 3.4 two hobos

2014-09-12T11:58:20Z

problem 3.4 two hobos

2014-09-12T11:57:42Z

problem 3.4 two hobos

2014-09-12T11:53:24Z

problem 3.4 two hobos

2014-09-12T11:50:02Z

problem 3.4 two hobos

@@ Line 39: / Line 39: @@
 ;Conservation of momentum
 :<math>0 = m(u-v_1) - (m+M) v_1</math>
-:<math>v_1 = \frac{m}{M+2m}u</math>: same as before
+:<math>v_1 = \frac{m}{M+2m}u</math>: no factor of 2 like before
 But now the second hobo jumps off

← Older revision		Revision as of 12:06, 12 September 2014
Line 1:		Line 1:
	An Inelastic collision conserves Momentum But Not energy		An Inelastic collision conserves Momentum But Not energy
		+
		+	:<math>\vec{P}_{\mbox{initial}} = \vec{P}_{\mbox{final}}</math>
		+

	Consider a collision between two bodies of mass <math>m_1</math> and <math>m_2</math> initially moving at speeds <math>v_1</math> and <math>v_2</math> respectively. They stick together after they collide so that each is moving at the same velocity <math>v</math> after the collision.		Consider a collision between two bodies of mass <math>m_1</math> and <math>m_2</math> initially moving at speeds <math>v_1</math> and <math>v_2</math> respectively. They stick together after they collide so that each is moving at the same velocity <math>v</math> after the collision.

← Older revision		Revision as of 12:01, 12 September 2014
Line 1:		Line 1:
−	An Inelastic collision ~~conservers~~ Momentum But Not energy	+	An Inelastic collision conserves Momentum But Not energy

	Consider a collision between two bodies of mass <math>m_1</math> and <math>m_2</math> initially moving at speeds <math>v_1</math> and <math>v_2</math> respectively. They stick together after they collide so that each is moving at the same velocity <math>v</math> after the collision.		Consider a collision between two bodies of mass <math>m_1</math> and <math>m_2</math> initially moving at speeds <math>v_1</math> and <math>v_2</math> respectively. They stick together after they collide so that each is moving at the same velocity <math>v</math> after the collision.

@@ Line 8: / Line 8: @@
 :<math>m_1 v_1 + m_2 v_2 = \left (m_1 + m_2 \right ) v</math>
-Given that the amsses and initially velocities are known we can solve for v such that
+Given that the masses and initially velocities are known we can solve for <math>v</math> such that
 :<math>v= \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2}</math>

@@ Line 12: / Line 12: @@
 :<math>v= \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2}</math>
 = problem 3.4 two hobos=
@@ Line 54: / Line 53: @@
 The flatcar will be moving faster if the hobos jump off separately.

← Older revision		Revision as of 11:59, 12 September 2014
Line 52:		Line 52:

	: <math> v_2= \frac{2M+3m}{2M+2m}\frac{2m}{M+2m}u =\frac{2M+3m}{2M+2m}v</math>		: <math> v_2= \frac{2M+3m}{2M+2m}\frac{2m}{M+2m}u =\frac{2M+3m}{2M+2m}v</math>
		+
		+	The flatcar will be moving faster if the hobos jump off separately.

← Older revision		Revision as of 11:58, 12 September 2014
Line 51:		Line 51:
	if written in terms of the velocity from part (a)		if written in terms of the velocity from part (a)

−	: <math> v_2= \frac{2M+3m}{2M+2m}\frac{2m}{M+2m}u = \frac{2m}{M+2m}v</math>	+	: <math> v_2= \frac{2M+3m}{2M+2m}\frac{2m}{M+2m}u =\frac{2M+3m}{2M+2m}v</math>

← Older revision		Revision as of 11:57, 12 September 2014
Line 48:		Line 48:
	: <math>(M+m) v_2= \frac{m(2M+3m)}{M+2m}u</math>		: <math>(M+m) v_2= \frac{m(2M+3m)}{M+2m}u</math>
	: <math> v_2= \frac{m(2M+3m)}{(M+2m)(M+m)}u</math>		: <math> v_2= \frac{m(2M+3m)}{(M+2m)(M+m)}u</math>
		+
		+	if written in terms of the velocity from part (a)
		+
		+	: <math> v_2= \frac{2M+3m}{2M+2m}\frac{2m}{M+2m}u = \frac{2m}{M+2m}v</math>

@@ Line 37: / Line 37: @@
 ;Conservation of momentum
 :<math>0 = m(u-v_1) - (m+M) v_1</math>
-:<math>v_1 = \frac{2m}{M+2m}u</math>: same as before
+:<math>v_1 = \frac{m}{M+2m}u</math>: same as before
 But now the second hobo jumps off
 :<math>m(u-v_1) - (m+M) v_1 = m(u-v_1) + m(u-v_2) - (M) v_2</math>
-: <math>-(m+M)\frac{2m}{M+2m}u=  mu- (M+m) v_2</math>
+: <math>-(m+M)\frac{m}{M+2m}u=  mu- (M+m) v_2</math>
-: <math>(M+m) v_2=  mu + (m+M)\frac{2m}{M+2m}u</math>
+: <math>(M+m) v_2=  mu + (m+M)\frac{m}{M+2m}u</math>
-: <math>(M+m) v_2=  \frac{m(M+2m)+2m(m+M)}{M+2m}u</math>
+: <math>(M+m) v_2=  \frac{m(M+2m)+m(m+M)}{M+2m}u</math>