A small wooden block with mass 0.800 kg is suspended from the lower end of a lig

Question

A small wooden block with mass 0.800 kg is suspended from the lower end of a light cord that is 1.60 m long. The block is initially at rest. A bullet with mass 0.012 kg is fired at the block with a horizontal velocity V­0 . The bullet strikes the block and becomes embedded in it. After the collision the combined object swings on the end of the cord. When the block has risen a vertical height of 0.800 m, the tension in the cord is 4.80 N. What is the initial velocity of the bullet?

(Got 281.43 m/s, but i am afraid its is wrong) thank you in advance

Explanation / Answer

Let the velocity of the system at the instant the tension in the cord is 4.90N be v.

also, 1.6* (1-cos Theta )=0.8

Theta = 60 deg deg

Now use Net tension T =mgcos theta + mv2/r

4.8 =0.812 *9.8cos 60 +0.812*v2/1.6

v=1.618 m/s

Now agian from Conservation of Energy

Kinetic energy of system just after collision= gain in potential energy+kinetic energy at height 0.725 m

=0.812*9.8*0.8 +0.5*0.812 *1.618*1.618 = 7.42 J

velocity at bottom= V^2 = 2 *KE/m = 2* 7.42/0.812

V = 4.27 m/s

finaly apply Conservation of momentum principle

mV=(m+M)v

0.012 V=0.812 *4.27

V= 288.93 m/s oe 289 m/s

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