Answer to Question #187235 in Mechanics | Relativity for Kanmi Matthews

Question

A particle, P, of mass 4kg was released with an initial velocity of 4m/s from a height of 80m above the horizontal ground. At the same time, another particle Q of mass 2kg was thrown vertically upwards with a velocity of 30m/s from the same horizontal ground directly below P. If the particles collide at a distance of 5m from the ground, calculate, correct to one decimal place, the 

i) time when P and Q collided

ii) velocities of the particles at the point of collision. 

iii) common velocity after the collision, if they moved together after collision.

Answer 1

Solution.

$m_P=4kg;$

$v_{0P}=4m/s;$

$h_{0P}=80m;$

$m_Q=2kg;$

$v_{0Q}=30m/s;$

$h_{0Q}=0m;$

$h=5m;$

$i) t-?;$

$h=v_{0Q}t-\dfrac{gt^2}{2};$

$5=30t-\dfrac{9.8t^2}{2};$

$4.9t^2-30t+5=0;$

$D=900-98=802;$

$t=\dfrac{30+\sqrt{802}}{2\sdot4.9}=6s;$ The particle will be at this height when it is already falling down.

$t=\dfrac{30-\sqrt{802}}{2\sdot4.9}=0.2s;$

$ii) v_Q=v_{0Q}-gt;$

$v_Q=30m/s-9.8m/s^2\sdot0.2s=10.4m/s;$

$v_P=v_{0P}+gt;$

$v_P=4m/s+9.8m/s^2\sdot0.2s=6m/s;$

$iii) v_{P}m_P-v_{Q}m_Q=(m_Q+m_P)v;$

$v=\dfrac{v_{P}m_P-v_{Q}m_Q}{m_Q+m_P};$

$v=\dfrac{6m/s\sdot 4kg-10.4m/s\sdot2kg}{2kg+4kg}=0.5m/s$ ;

Answer: $i) t=0.2s;$

$ii)v_Q=10.4m/s; v_P=6m/s;$

$iii)v=0.5m/s.$