Question

Two blocks of masses

*m*_{1} = 1.50 kg

and

*m*_{2} = 3.00 kg

are each released from rest at a height of

*h* = 4.40 m

on a frictionless track, as shown in the figure below, and undergo an elastic head-on collision. (Let the positive direction point to the right. Indicate the direction with the sign of your answer.)

(a) Determine the velocity of each block just before the collision.

(b) Determine the velocity of each block immediately after the collision.

(c) Determine the maximum heights to which

*m*_{1}

and

*m*_{2}

rise after the collision.

Answer #1

**a)** both masses are released from rest

by energy conservation

change in potential energy = change in kinetic energy

mgh = 1/2mv^{2}

so v= sqrt(2gh) where sqrt = squar root

for mass1 m1= 1.5kg

V_{1} = sqrt(2*9.8*4.40) = **9.29
m/sec**

for mass M_{2}

V_{2} = sqrt(2*9.8*4.40) = **9.29
m/sec**

b) elastic collision both conservation of momentum and kinetic energy are observed

so velocity of M_{1} after collision (V _{3}) =
5*9.29/3 = **15.83 m/sec**

and velocity of M2 after collision (V _{4}) =
9.29/3 = **3.1 m/sec**

**c)** again energy conservation

1/2 mv^{2} = mgh

h = v^{2} /2g

so mass 1 H_{1} = 15.83^{2} /2*9.8 =
**12.78 m**

and mass 2 H_{2} = 3.1^{2} / 2*9.8 =
**0.49 meter**

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