A 0.5 kg block, block-A, sliding frictionless in the x-direction at an initial speed of 1.5 m/s, collides with a stationary block, block-B, of mass m . The collision is elastic. After the collision, block-A continues traveling in its original direction at 0.7 m/s. (a) What is the mass of the second block? (b) What is its speed after the collision ? (c) What is the speed of the center of mass of the two blocks? Note: The velocity of the CM for a system of particles is calculated similarly to the position of the CM. (d) If the block-A is in contact with block-B for 0.2 ms, what is the average force applied on block-B? (e) What is the impulse?​

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the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s

the maxium height is 34m/s