# Math Help - Kinematics

1. ## 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!

2. 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 $a = g\sin{\theta}$

basic kinematics equation ...

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

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

motion of particle S up the incline ...

$x = 8t - 2.45t^2$

motion of particle T down the incline ...

$x = 4 - 2.45t^2$

solve the system for $x$ , the position relative to point P where the collision occurs.

3. thanks for the respond.

can i just ask where did the 2.45 come from?

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

ok thanks!

4. 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) = $\frac{13}{60}$

So for d) i did:
$
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?