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Kobotan [32]
3 years ago
7

He shoots the tree stump, which has a mass of (M), with a bullet of mass (m) traveling at some velocity (vbullet), and the bulle

t lodges itself inside the tree stump before the system of objects slides along the ice. What is the speed of the tree stump-bullet system immediately after the collision?
Physics
1 answer:
Ymorist [56]3 years ago
4 0

Answer:

Explanation:

Given

mass of tree stump is M

mass bullet is m

velocity of bullet is v

Conserving momentum for bullet and tree stump

Initial Momentum P_i=mv

Suppose v_0 is the velocity of the system

Final Momentum P_f=(M+m)v_0

Initial momentum =Final Momentum

mv=(m+M)v_0

v_0=\frac{mv}{m+M}

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Mass m moves to the right with speed =v along a frictionless horizontal surface and crashes into an equal mass m initially at re
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After the collision the magnitude of the momentum of the system is Mv

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mass of 1st object = M

speed of 1st object = v

mass of 2nd object = M

speed of 2nd object = 0

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magnitude of the momentum after collision

Solution: Product of the mass of a particle and its velocity. Momentum is a vector quantity; i.e., it has both magnitude and direction. Isaac Newton's second law of motion states that the time rate of change of momentum is equal to the force acting on the particle.

Applying conservation of linear momentum

Mv + M(0) = 2MV

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2 years ago
A 61 kg skater is traveling at 2.5 m/s while carrying a 4.0 kg bowling ball. After he throws the bowling ball forward at twice t
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The final velocity of the skater is 2.34 m/s forward

Explanation:

We can solve this problem by using the law of conservation of momentum. In fact, the total momentum of the system before and after the ball is thrown must be conserved, in absence of external forces.

Before the ball is thrown, the total momentum is:

p_i = (M+m)u

where

M = 61 kg is the mass of the skater

m = 4.0 kg is the mass of the ball

u = 2.5 m/s (forward) is the combined velocity of the skater and the ball

After, the ball is thrown at twice the velocity, so the final total momentum is

p_f = MV+mv

where

V is the final velocity of the skater

v = 2(2.5) = 5.0 is the final velocity of the ball

Since the total momentum must be conserved, we can write

p_i = p_f\\(M+m)u = MV+mv\\V=\frac{(M+m)u-mv}{M}=\frac{(61+4.0)(2.5)-(4.0)(5.0)}{61}=2.34 m/s

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