MHB Mechanics- connected particles

[Shah 72]

I got the speed of p when q reaches the pulleys = 1m/s, a= 2m/s^2
Iam getting time = 0.8s for q(b)

 

[skeeter]

You really need to figure out how to post a readable image ...

Both masses will have the same magnitude of acceleration. The tension in the string on both sides of the pulley will be the same.

$Mg - T = Ma$

$T - mg = ma$

solve for the system for acceleration, then use your kinematics equations for uniformly accelerated motion to answer the questions.

 

[Shah 72]

You really need to figure out how to post a readable image ...

View attachment 11169

Both masses will have the same magnitude of acceleration. The tension in the string on both sides of the pulley will be the same.

$Mg - T = Ma$

$T - mg = ma$

solve for the system for acceleration, then use your kinematics equations for uniformly accelerated motion to answer the questions.

I got the ans for this.

 

[Shah 72]

You really need to figure out how to post a readable image ...

View attachment 11169

Both masses will have the same magnitude of acceleration. The tension in the string on both sides of the pulley will be the same.

$Mg - T = Ma$

$T - mg = ma$

solve for the system for acceleration, then use your kinematics equations for uniformly accelerated motion to answer the questions.

Iam still having doubts with q(b)
For q(a)
I got a= 2m/s^2 and speed of p when q reaches the pulley = 1m/s
Q(b)Time when the system is released v= u+at1
t1=0.5s
After the string breaks, there is constant speed
I don't understand

 

[Shah 72]

You really need to figure out how to post a readable image ...

View attachment 11169

Both masses will have the same magnitude of acceleration. The tension in the string on both sides of the pulley will be the same.

$Mg - T = Ma$

$T - mg = ma$

solve for the system for acceleration, then use your kinematics equations for uniformly accelerated motion to answer the questions.

 

[skeeter]

when the string breaks, P is 1.2m above the ground moving downward with initial speed of 1 m/s and is in a state of free fall.

$\Delta y = v_{y_0} \cdot t_2 - \dfrac{1}{2}g t_2^2$

 

[Shah 72]

when the string breaks, P is 1.2m above the ground moving downward with initial speed of 1 m/s and is in a state of free fall.

$\Delta y = v_{y_0} \cdot t_2 - \dfrac{1}{2}g t_2^2$

The ans is 0.1s but the textbook says 0.9s

 

[Shah 72]

when the string breaks, P is 1.2m above the ground moving downward with initial speed of 1 m/s and is in a state of free fall.

$\Delta y = v_{y_0} \cdot t_2 - \dfrac{1}{2}g t_2^2$

Oh I got it. It will be a quadratic equation and I solve using quadratic formula

 

[Shah 72]

when the string breaks, P is 1.2m above the ground moving downward with initial speed of 1 m/s and is in a state of free fall.

$\Delta y = v_{y_0} \cdot t_2 - \dfrac{1}{2}g t_2^2$

Thank you so much. t2= 0.4s so total time will be 0.9 s