# Kinematics

• Nov 27th 2009, 04:56 AM
BabyMilo
Kinematics
P and Q are 4m apart at opposite ends of a smooth plane inclined at 30 degree to the horizontal. Initially a particle S of mass m is projected up the plane from P with speed 8m/s and a particle T of mass 2m is released from rest at Q.

Find the distance of the particles from P at the instant when they collide.

Thanks!
• Nov 27th 2009, 06:06 AM
skeeter
Quote:

Originally Posted by BabyMilo
P and Q are 4m apart at opposite ends of a smooth plane inclined at 30 degree to the horizontal. Initially a particle S of mass m is projected up the plane from P with speed 8m/s and a particle T of mass 2m is released from rest at Q.

Find the distance of the particles from P at the instant when they collide.

Thanks!

magnitude of acceleration a smooth incline is $\displaystyle a = g\sin{\theta}$

basic kinematics equation ...

$\displaystyle x = x_0 + v_0 t + \frac{1}{2}at^2$

let point P be $\displaystyle x_0$ , and up the incline be the (+) direction.

motion of particle S up the incline ...

$\displaystyle x = 8t - 2.45t^2$

motion of particle T down the incline ...

$\displaystyle x = 4 - 2.45t^2$

solve the system for $\displaystyle x$ , the position relative to point P where the collision occurs.
• Nov 27th 2009, 06:55 AM
BabyMilo
thanks for the respond.

can i just ask where did the 2.45 come from?

oh i see. $\displaystyle \frac{1}{2} * g * sin30$

ok thanks!
• Nov 27th 2009, 07:22 AM
BabyMilo
d) Find the length of the time that elapses from the beginning of the motion until the combined particle reaches P.

for a,b,c I found out:

the distance of the particles from P = 3.3875m

Vs = 5.55ms^-1

Vt = 2.45ms^-1

Vs+t(after collision) = $\displaystyle \frac{13}{60}$

So for d) i did:
$\displaystyle 3.3875=\frac{13}{60}t+\frac{1}{2}g sin 30*t^2$

which t=1.13s (3.s.f)

the answer therefore is 1.13+0.5 = 1.63s

but the actual answer is 1.816 s

where have i gone wrong?