An object A moving along the x axis with a velocity v undergoes an elastic colli

Question

An object A moving along the x axis with a velocity v undergoes an elastic collision with an identical object B that is initially at rest (as shown). After the collision, A is observed to move along a direction making an angle of 30 degree with the x axis, while B is moving along a direction as shown in the diagram. What is the value of the angle theta? a. 60 degree b. 45 degree c. 30 degree d. 90 degree (For no. 7 and no. 8) A "ballistic" pendulum is set up to measure the sped of a bullet. It consists of a heavy block of mass M = 2.50 kg hanging vertically by a light wire. A gun is fired horizontally with the bullet entering with a high velocity v and embedded in the block. After the embedding, the whole system (block and bullet) is observed to rise to a maximum height of h = 25.0 cm. If the mass of the bullet is only 2.50 g (grams), then The velocity v' of the system (block and bullet) with which it starts to swing upward is equal to: a. 0.700 m/s b. 7.00 m/s c. 2.21 m/s d. 1.70 m/s The velocity v of the bullet can then be obtained as given by: a. 7.00 times 10^2 m/s b. 3.50 times 10^3 m/s c. 1.71 times 10^3 m/s c. 2.22 times 10^3 m/s

Explanation / Answer

7. Applying energy conservation after the collision,

total energy just after collision = total energy at highest point

{ total energy = PE+ KE}

0 + m v^2 /2 = 0 + m g h

v = sqrt(2 x 9.8 x 0.25) = 2.21 m/s ........Ans

Ans(c)

8.
Applying momentum conservation for the collison,

m v0 + Mx 0 = (m + M)v

(2.50 x 10^-3) v0 = ((2.50 x 10^-3) + 2.50) (2.21)

v0 = 2.22 x 10^3 m/s

Ans(c)

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