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 r-soy June 12th, 2012 06:54 AM

Direct central impact occurs between a 300N body moving to

Hi all

1 )
Direct central impact occurs between a 300N body moving to the right with velocity of 6 m/s and 150 N body moving to the left with velocity of 10 m/s . Find the velocity of each body after impact if the coefficient of restitution is 0.8 .

[color=#BF4000]Given [/color]::
m1 = 300 /9.81 = 30.58 kg
u1 = 6 m/s
m2 = 150/9.81 = 15.2 kg
u2 = -10 m/s
e = 0.8
V 1 ? v2 ?

m1u1+m2u2 = m1v1+m2v2
30.58X 6 + 15.2 X (-10) = 30.58v1+ 15.2 v2
183.48 -152 = 30.58v1+ 15.2 v2
31.48 = 30.58v1+ 15.2 v2 ===== [color=#FF0000]this is first equation (1)[/color]

e(u1 - u2) = (v2 - v1)
0.8(6 + 10) = v2 - v1
4.8 + 8 = v2 - v1
12.8 = v2 - v1
v2 = 12.8 + v1 ===== [color=#FF0000]this is first equation (2) [/color]

make v2 in fist equation
31.48 = 30.58v1+ 15.2 (12.8 + v1)
31.48 = 30.58v1+ 194.56 + 15.2 v1)
31.48 - 194.56= 30.58v1 + 15.2 v1
-163.08 = 45.78
[color=#0040FF]v1 = -3.6 m/s [/color]

[color=#0040BF]to find v2 = 12.8 + (-3.6) = 9.2 m/s[/color]

 MarkFL June 13th, 2012 09:27 AM

Re: Direct central impact occurs between a 300N body moving

Let the body initially moving to the right be body A and the other be body B, hence:

$v_a=\frac{m_a u_a+m_b u_b+m_bC_R$$u_b-u_a$$}{m_a+m_b}$

$v_b=\frac{m_a u_a+m_bu_b+m_aC_R(u_a-u_b)}{m_a+m_b}$

We are given:

$m_a=\frac{300}{9.81}\:\text{kg}=\frac{10000}{327}\ :\text{kg}$

$u_a=6\text{ \frac{m}{s}}$

$m_b=\frac{150}{9.81}\:\text{kg}=\frac{5000}{327}\: \text{kg}$

$u_b=-10\text{ \frac{m}{s}}$

$C_R=0.8$

Hence:

$v_a=-3.6\text{ \frac{m}{s}}$

$v_b=9.2\text{ \frac{m}{s}}$

 r-soy June 14th, 2012 05:38 AM

Re: Direct central impact occurs between a 300N body moving

that my answer is correct same which you are got

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