oooo Verizon 9:50 PM 40% A bullet 0250 kg) is firéd with a speed of 97.00 m/s an

Question

oooo Verizon 9:50 PM 40% A bullet 0250 kg) is firéd with a speed of 97.00 m/s and hits a block (M = 2.3 kg) supported by two light strings as shown, m stopping ind the height to whích the block rises 4.795 m The bullet stops (quickly) due to an INTERNAL force, ie, that between the wood and the bullet. The horizontal momentum of the two is conserved in the collision and then changes to zero as the tension acts on the System Mechanical energy is conserved during the swing of the block. Check your units. Submit Answer ) Incorrect. Tries 3/5 Submit Answer Incorre Previous ious ies

Explanation / Answer

Here

m = 0.0250 Kg

speed of bullet , u = 97 m/s

M = 2.3 Kg

let the speed of block just after collision is v

Using conservation of momentum

(m + M) * v = m * u

(0.0250 + 2.3) * v = 0.0250 * 97

solving for v

v = 1.011 m/s

Now , for the maximum height

maximum height = v^2/(2 * 9.8)

maximum height = 1.011^2/(2 * 9.8)

maximum height = 0.0521 m

the height to which the block rises is 0.0521 m
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let the angle is theta

h = L ( 1 - cos(theta))

0.0521 = 0.350* (1 - cos(theta))

solving for theta

theta = 31.7 degree

the angle through which the block rises is 31.7 degree

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